Transcript
Page 1: Soal Fisika Dan Jawaban

SOAL SOAL SOAL SOAL

Di Buat Oleh :

- Abdul Rahman Halim

- Alfi Rahmah Santika

- Amalia Hajar D.E.

- Anis Fitriyah

- Annisah Maharinda

- Muhammad Fiqhi Ibad

- Muhammad Rizqi Alfian

- Nanda Pradana

- Muhammad Fatkhul Arifian

- Nandini candrika

- Zumrotul Ulya

- Yuli Lestari

- Razan Miqdad Afifi

- Azizah Ardyanti Putri

- Azizah Handayani

- Elma Elviana

SOAL SOAL SOAL SOAL –––– SOAL DAN JAWABANSOAL DAN JAWABANSOAL DAN JAWABANSOAL DAN JAWABAN

““““FISIKA (PHYSICS)FISIKA (PHYSICS)FISIKA (PHYSICS)FISIKA (PHYSICS)””””

- Dimas Fasihatin

- Inge Nindiana Irawan

- Muhimmatus Syarifah

- Veronika Prastiwi

- Tika Aprilia

- Muhammad Ainul Yaqin

Muhammad Rizqi Alfian - Clarin Puspa

- Devi Ratna Safitri

Muhammad Fatkhul Arifian - Asri Lindung Sari

- Muhammad Tulus Subakti

SMA NAHDLATUL ULAMA 1 GRESIK

TAPEL 2010

Dimas Fasihatin

Inge Nindiana Irawan

Muhimmatus Syarifah

Veronika Prastiwi

Muhammad Ainul Yaqin

Devi Ratna Safitri

Asri Lindung Sari

Muhammad Tulus Subakti

SMA NAHDLATUL ULAMA 1 GRESIK

TAPEL 2010 - 2011

Page 2: Soal Fisika Dan Jawaban

ALAT OPTIK

1. Rani who suffer from nearsightedness has a focal point 50 cm. If you want to read with normal spacing (25 cm), then what is the point nearby? Diketahui : s = 25 cm f = 50 cm Ditanyakan: s’ = ... ? Jawab :

1

�=

1

�+

1

�′

1

50=

1

25+

1

�′

1

�′=

1

25−

1

50

1

�′=

2

50−

1

50

s’ = 50cm = 0,5m

2. In experiments on optics, John uses convex lens that focuses 25 cm. In front of the lens is placed objects whose height 3 cm at a distance of 30 cm. Specify: a. shadow distance to the lens, b. magnification image, c. high shadow? Completion f = + 25 cm (cembung) S = 30 cm dan h = 3 cm a. Jarak bayangan ke lensa S’ memenuhi:

1

�=

1

�+

1

�′

1

25=

1

30+

1

�′

1

�′=

1

25−

1

30=

6 − 5

150=

1

150

Berarti S = 150 cm b. Perbesaran bayangan sebesar :

M =�′

�=

150

30= 5x(nyata)

c. Tinggi bayangan benda sebesar:

M =ℎ′

h’=M.h = 5 . 3 = 15cm

Page 3: Soal Fisika Dan Jawaban

SUHU DAN KALOR KD 4.1

• The result of water temperature measurements using the thermometer scale Celsius is 40 º C. Determine the number that is shown if the water temperature was measured with a thermometer Kelvin scale!

Answer: Diket : C = 40º C

100=

� − 273

100

40

100=

� − 273

100

K = ̶ 273 – 40 = ̶ 313ºK

• When the ice is melting, the thermometer scale showed the 20 º X X and while it is boiling, the thermometer scale showed the 140 º X X. This situation was measured at 1 atmosphere air pressure. Find The relationship between scale C and scale X Thermometer scale of C, if the thermometer scale showed 80 º X X!

Answer:

Diket : Pembagian skala X = (140-20) skala = 120 skala C : X = 100 : 120 = 5 : 6

a. Hubungan skala C dan X. �

5=

� − 20

6

b. 80 ºX = ... ºC �

5=

(80 − 20)

6

� =60 × 5

6= 50°�

Jadi, skala C menunjukkan angka 50ºC

• If the scale Fahrenheit thermometer showed a temperature of 50 º F. Find the scale indicated by the

Celsius and Kelvin!

Answer:

Diket : F = 50ºF

� − 32

180=

100

!"#$

%&!=

'

%!!

%&

%&!× 100 = �

C = 10ºC

Page 4: Soal Fisika Dan Jawaban

� − 32

180=

� − 273

100

!"#$

%&!=

("$)#

%!!

%&

%&!× 100 = � − 273

10 = � − 273

10 + 273 = K

K = 283 ºK

• A rectangular aluminum plate with the long side - side 50 cm and temperature 30 º C. If the coefficient

length of aluminum miaui 25. 10-6 º C-1, determine the aluminum square area if the temperature is raised

to 150 º C!

Answer:

� − 32

180=

100

!"#$

%&!=

'

%!!

%&

%&!=

'

%!!

C = 10 ºC

� − 32

180=

� − 273

100

%&

%&!=

("$)#

%!!

10 = K – 273

10 + 273 = K

K = 283 ºK

• A bullet-shaped ball made of copper with a radius of 3 mm at a temperature of 20 º C. If the length expansion coefficient of copper 17. 10-6 º C-1, determine the volume of bullets if heated to 170 º C!

Answer:

Diket ro = 3mm; α = 17 . 10-6 ºC-1 To = 20 ºC; T1 = 170 ºC Vt = Vo (1+γ∆T) dengan γ = 3α = (51 . 10-6 ºC-1) ∆+= (170 -20) ºC = 150 ºC

,o = -

#./0

3 = -

#(3,140(3 22)3 = 113,04 mm3

� Sehingga didapatkan Vt = 113,04 mm3 {1 + (51 . 10-6 ºC-1-(150 ºC)} = 113,04 (1 + 0,00765)mm3 Vt = 113,04 (1,00765) = 113,90 mm3

Page 5: Soal Fisika Dan Jawaban

• A gas when the pressure is 6 atmospheres volume 2 liters. For fixed temperature, if pressure is reduced to 4 atmosphere, calculate the volume occupied by gas! (1 atm = 1.013. 105 Pa)

Answer:

V l = 2 L, P1 = 6 atm, dan P2 = 4 atm P1 v1 = p2 v2

( 6 atm)( 2 L) = ( 4 atm )V2

,$ =6 × 2

4 3 = 3 3

• A 4 meter long iron rod with cross-sectional area 25 cm2. The difference in temperature between both ends of the iron 80 º C. Also known coefficient of thermal conductivity of iron 46 Jm-1 º C-1. Determine the amount of heat that crept in the iron per unit of time!

Answer:

l = 4 m; ∆+= (T2 – T1) = 80 ºC

A = 25 cm2 = 25 . 10-4m2 k = 46 Jm-1 ºs-1 ºC-1

4 = 56∆+

7

H = (46 Jm-1s-1 ºC-1)(25 . 10-4 m2)&!°8-9 = 2,3 :�"% = 2,3 ;<==

KD 4.2

• Determine the amount of heat needed to heat the water as much as 2 kg of a temperature of 20 º C until it reaches its boiling point of 100 º C. known to heat water type kJkg 4.2 º C-1!

Answer:

m = 2 kg, c = 4,2 kJkg-1ºC-1 ∆T = 100 ºC - 20 ºC = 80 ºC > = 2 . @ . ∆+ = (2kg)(4,2 kJ kg-1 ºC-1)(80 ºC) = 672 kJ

• An object has a heat capacity of 2000 J ° C-1. Determine the amount of heat required to raise the body temperature of 10 º C

Answer:

Diket : C = 2.000 JºC-1 ∆T = 10 ºC Q = C∆+= (2.000 JºC-1)(10 ºC) = 20.000 joule = 20kJ

Page 6: Soal Fisika Dan Jawaban

KD 4.3

• In a cup of tea there are as many as 60 mL of water with a temperature of 80 º C. Then, into the glass was added to 40mL of water temperature of 5 º C. If known heat tea water type with heat cold water species, determine the temperature of the mixture of water!

Answer:

Diket : mteh = 60mL, Tteh = 80 ºC mair = 40mL, Tair = 5 ºC cteh = cair jika suhu akhir camouran adalah Tc, diperoleh Qlepas (teh) = Qterima(air es)

mteh cteh ( Tteh – Tc ) = mair cair ( Tc – Tair ) 60 mL ( 80 ºC – Tc ) = 40mL ( Tc - 5 ºC ) 4800 ºC – (60Tc) = ( 40 Tc ) - 200 ºC 100 Tc = 5.000 ºC Tc = 50 ºC

• A metal type X is determined heat put into a kalorima = ether. X metal mass is 0.2 kg. Temperature at first - first 98 º C. Known also Caloriemeter made of polystyrene material, which material can absorb almost no heat. In the calorimeter contained as much as 1 kg of water temperature 25 º C. After X Included metal into the calorimeter, then stirred, thermal equilibrium is reached a temperature of 28 º C. From these observations, determine the heat of the metal type X, if you heat water type Jkg 4200 º C-1! Answer:

ml cl (Tl – Tt) = maca ( Tt – Ta )

(0,2 kg)cl (98 – 28) ºC = (1 kg)(4.200 Jkg-1 ºC-1)(28 – 25) ºC

cl =

�-.$!!�×�#��!,$�×�)!� :5A-1 ºC-1 = 900 Jkg-1 ºC-1

• Know the melting heat of water 3.36. 105Jkg-1 and m = 0.1 kg, Determine the amount of heat required to melt 100 grams of ice! Answer: Diket : L = 3,36 . 105 Jkg-1 dan m = 0,1 kg

Q = m . L = (0,1 kg)(3,36 . 105 Jkg-1) = 3,36 . 104 J

• Known heat frozen water 3.36 Jkg -1. Determine the amount of water which freezes when the temperature is 0 º C absorbed energy of 1.68. 104 J of water.

Answer:

Q = m . L

m =BC = %,D& × %!E

#,#D×%!FGHIJK = 0,05 5A = 500A/<2

Page 7: Soal Fisika Dan Jawaban

• By using steam heat, determine the amount of heat required to evaporate 0.5 kg of water at a temperature of 100 º C so that it becomes steam at 100 º C!

Answer: Q = m . L = (0,5 kg)(2,26 . 106Jkg-1) = 1,13 . 106 Joule • Diketahui 400 gram es bersuhun 0 ºC diberikan kalor sebesar 3,36 kJ. Jika kalor lebur es 3,36 . 105Jkg-1, tentukanlah berapa persen es yang melebur !

Jawab : Q = m . L

m =#,#D HG

##DGHIJK = 0,01 5A; MN OPQR SMTMUVW = !,!% HI!,- HI × 100% = = 25%

• Determine the amount of heat required to raise the temperature of 0.2 kg of ice from -10 º C until all the water temperature of 50 º C 1 is known heat melting ice 3.36. Jkg 105-1, heat type of ice Jkg 2100 º C-1- 1, and heat water type Jkg 4200 º C-1-1.

Answer: Diket : mes = 0,2 kg; Les = 3,36 . 105 Jkg-1

To = -10 ºC; ces = 2.100 Jkg-1 ºC-1 T1 = 4200 ºC; cair = 4.200 Jkg-1 ºC-1

Q = Q1 + Q2 + Q3 Q1 = mes ces ∆es =(0,2 kg)(2.100 Jkg-1 ºC-1)(0-(-10)) ºC = 4,2 kJ Q2 = mes Les = ( 0,2 kg)(3,36 . 105 Jkg-1) = 67,2 kJ Q3 = mair cair ∆air = (0,2 kg )(4.200 Jkg-1 ºC-1)(50 – 0) ºC = 42 kJ Qtotal = Q1 + Q2 +Q3 = 113,4 kJ

• The length of a rod of iron at a temperature of 20 º C is 10 meters. If the length expansion coefficient of

iron 11. 10-6 º C-1, determine the length of iron bar was added if the temperature is raised to 100 º C!

Answer:

Diket : To = 20 ºC; T1besi = 100 ºC

lo = 10 m ; α = 11 . 10-6 ºC-1

∆l

= αlo ( T1 – To )=(11 . 10-6 ºC-1)(10 m )(100 – 20)ºC= 8,8 . 10-3meter

∆l = 8,8 milimeter

( ( ( ( ABDURRAHMAN HALIMABDURRAHMAN HALIMABDURRAHMAN HALIMABDURRAHMAN HALIM ))))

Page 8: Soal Fisika Dan Jawaban

KD 4.1

1. At 30 ° C a metal plate breadth of 10 m2. If the temperature raised to 90 ° C and length expansion coefficient of iron 0.000012 / ° C, determine the area of iron plate!

Diketahui : A1 = 10 m2

T1 = 30° C T2 = 90° C T = T2 – T1 = 90 – 30 = 60° C

= 0,000012/° C = 2 × 0,000012/° C = 0,000024

Ditanyakan : A2 = ... ? Jawab : A2 =A1(1 +β x ∆T) = 10(1 + 0,000024 × 60) = 10(1 + 0,00144)

= 10 × 1,00144 = 10,0144 m2

Jadi, luas pelat besi setelah dipanaskan adalah 10,0144 m2.

2. How much heat is required to convert 2 grams of ice at 0 ° C to water vapor at a temperature of 100 ° C? (Liquid = 4200 J / kg ° C, L = 336J/Kg, and U = 2260 J / Kg)

Diketahui : m = 2 g = 2 × 10-3 kg

T = 100° – 0° = 100° C U= 2.260 J/Kg L = 336J/Kg Cair = 4.200 J/kg °C

Ditanya : Qtotal = ……? Dijawab : Q1 = m x L = 2 x 336 J/Kg

= 672 J Q2 = m x Cair x ∆T = 2.10-3 x 4200 x 100 = 840 J Q3 = m x U = 2 x 2260 = 4420 J Qtotal = Q1 + Q2 + Q3

= 672 + 840 + 4420 = 5932 J

Page 9: Soal Fisika Dan Jawaban

3. A piece of iron which has a mass of 3 kg, is heated from temperature 20 ° C to 120 ° C. If heat is absorbed iron by 135 kJ. Determine the heat capacity of iron and heat type of iron?

Diketahui : m = 3 kg

T = 120° – 20° = 100° C Q = 135 kJ

Ditanya : C = …..? c = …..?

Dijawab : a. C = B

∆Y

= %# .!!!

%!!

= 1350 J/0C 4. Ethyl alcohol move at 78.50 C and frozen at-1170C padatekanan 1 atm. Express this temperature in both a. Kelvin b. Fahrenheit

Dijawab: K = C + 273 = 78,5 + 273 = 351,5 K K = -117+ 273 = 156 K F = 9/5C + 32 = 9/5(78,5) + 32 = 1730F F = 9/5(-117) + 32 = -1790F

Page 10: Soal Fisika Dan Jawaban

5. A container made of aluminiumyang filled with 300mL of glycerin at a temperature of 200C. What is the amount of glycerin that tumpaah if the container is heated to suhu1100C? Length expansion coefficient of aluminum 2.55 x 10-5/0C and volume expansion coefficient of 5.3 x 10-4/0C glycerin.

γtampak = γfluida - γwadah

= 5,3 x 10 – 3(2,55 x 10-5)

= 4, 535 x 10-4

∆V = V0 . γ . ∆T = 300 x 4, 535 x 10-4 (110 – 20 )

= 12,24 mL

6. One object with kalorrs capacity 1500 Joule/0C. How much calor one at need to raise that object temperature as big as 400C

C = 1500 Joule/0C ∆T = 400C Q = C ∆T 1500 . 40 6000 Joule

KD 4.3

1. As much as 0.5 kg of water with a temperature of 100 ° C in pour into vessel of aluminum which has a mass of 0.5 kg. If the initial temperature vessel at 25 ° C, heat of aluminum 900 J / kg ° C, and heat types of water 4200 J / kg ° C, determine the equilibrium temperature achieved! (assume no heat is flowing into the environment)

Diketahui : mbjn = 0,5 kg mair = 0,5 kg Tair = 100° C Tbjn = 25° C Cair = 4.200 J/kg °C Cbjn = 900 J/kg °C

Ditanyakan : Ttermal = ...? Jawab : Qdilepas = Qditerima m × Cair × Tair = m × Cbjn × Tbjn 0,5 × 4.200 × (100 – Ttermal) = 0,5 × 900 × (Ttermal – 25)

Page 11: Soal Fisika Dan Jawaban

210.000 – 2.100 Ttermal = 450 Ttermal – 11.250 2.550 Ttermal = 222.250

Ttermal = 222.250 2550 = 87,156° C Jadi, suhu kesetimbangannya adalah 87,156° C.

2. The water 150 grams in temperature 200C mixed with the water 100 grams in the temperature 900C. Determine the mixture temperature of both water. Cwater = 1 cal/g. Diketahui: m1 = 150 grams

m2

T1

T2

Cwater

Ditanyakaan: Tc= ……? Dijawab: Qlepas = Qterima

m C ∆T = m C ∆T 100.1(100 -Tc) = 150.1(Tc – 30) 100 (100 - TC) = 150 (Tc – 30) 10.000 – 100Tc = 150Tc – 4500 14.500 = 250Tc

%-. !!

$ ! = Tc 58 = Tc

3. The mass 10g ice with the temperature -10oC mixed the mass 20g water with temperature 80oC . If Cice = 0,5 cal/goC , cwater = 1 kal/goC and Lice = 80 cal/g , measure the mixture temperature .

Answer :

� D1 =m1 = mes = 10 g

m2 = mair = 20 g T1 = -10oC T2 = 80oC

� D2 Qobsorb = m1 . cice (0-(-T1)) + m1 Lice 10 . 0,5 (10) + 10 (80) 850 cal

Qrelease = m2 (T2 – 0) 20 (80 – 0) 1600 cal

Page 12: Soal Fisika Dan Jawaban

Qtot = m1 Cice {0- (-T1)} + m1 Lice +m1 (Tc – 0) = m2 (T2 – Tc) 10 . 0,5 {0-(-10)} + 10 . 80 +10Tc = 20 (80 – Tc) 50 + 800 +10Tc = 1600 – 20 Tc 50+800-1600 = -20Tc – 10Tc -750 = -30 Tc Tc = 250C

4. in a vessel of negligible mass contained 42 grams of water mixed with a gram of ice C -4 C. After the stirring was 50% ice melt. If titk melting ice = 0C ice type heat 0.5 cal / g C heat melting ice = 80 cal / gc, calculate the ratio of a and b

in accordance with the principle of black then the heat is released air (Q3) equal with calor absorbed to raise its temperature (Q1) and to melt ice 50% (Q2)

Q3 = Q1 + Q2 mwater Cwater ∆T

mice Cice ∆T + 0,5 mice L ice

a(1) 42 = b(0,5) 4 +0,5 (b) 80 42a = 42b atau a:b = 1:1

5. calculate the amount of heat required to melt 100 grams ice 150C in to water 500C, if Cwater = 4200J / Kg K , Cice = 2100 J/Kg K and Lice = 336.000 J/Kg .

Answer : � m = 100 gram = 0,1 kg

tice1 = 150C tice2 = 50oC Cwater = 4200J / Kg K Cice = 2100 J/Kg K L ice = 336.000 J/Kg .

� Qtotal ……….? � Q1 = m . c . ∆T

= 0,1 . 2100 . 15 = 3150 J

Q2 = m . L = 0,1 . 336.000 = 33600 J Q3 = m . c . ∆T = 0,1 . 4200 . 50 = 21.000 J

Page 13: Soal Fisika Dan Jawaban

Qtotal = Q1 + Q2 + Q3

= 57.750 J 6. calculate the amount of heat required to melt 50 grams ice 300C in to water 750C, if Cwater

= 4200J / Kg K , Cice = 2100 J/Kg K and Lice = 336.000 J/Kg . Answer :

� m = 50 gram = 0,01 kg tice1 = 300C tice2 = 75oC Cwater = 4200J / Kg K Cice = 2100 J/Kg K L ice = 336.000 J/Kg .

� Qtotal ……….? � Q1 = m . c . ∆T

= 0,05 . 2100 . 30 = 3150 J

Q2 = m . L = 0,05 . 336.000 = 16800 J Q3 = m . c . ∆T = 0,05 . 4200 . 75 = 31.500 Qtotal = Q1 + Q2 + Q3

=5 1.450 J

( ALFI RAHMAH SANTIKA )( ALFI RAHMAH SANTIKA )( ALFI RAHMAH SANTIKA )( ALFI RAHMAH SANTIKA )

Page 14: Soal Fisika Dan Jawaban

KD 4.1

1) The temperature of the liquid when measured with a thermometer centigrade scale shows the number 25°Z. What is the liquid temperature was measured with a thermometer:

a) the scale degrees reamur?

b) scale degrees Fahrenheit?

c) degrees kelvin scale?

→ Answer :

Data : t = 25 ℃

Problem : a) t °\ b) t ℉ c) T

a). t ℃ : t °\ = 5 : 4 → ^ ℃

^ °_ =

-

t °\ = -

t ℃ =

-

. 25 = 20

* So the temperature of a substance that shows 20°\

b). t ℃ : (t ℉ - 32) = 5 : 9

^ ℃

(^ ℉ " #$) =

`

t ℉ = `

t ℃ + 32

= `

. 20 + 32 = 77

* So the temperature of a substance that shows 77 ℉

c) t ℃ : (T – 273) = 5 : 5

^ ℃

(a"$)#) =

T = t ℃ + 273

= 25 + 273 = 298

* So the temperature of a substance that shows 298 ℉

2) What is the temperature of an object, if measured with a thermometer scale Celsius and Fahrenheit scales show the same number?

→ Answer :

Data : t ℃ = t ℉ = x

Page 15: Soal Fisika Dan Jawaban

Problem : x

a). t ℃ : (t ℉ - 32) = 5 : 9

9. t ℃ = 5 (t ℉ - 32)

a ℃

�^ ℃" #$� = `

9x = 5x – 160

9x – 5x = -160

X = - 40

* So the two thermometers will designate the number -40

3) What kind of heat a substance, if the mass of the object is equal to 100 grams of the temperature will rise by 80C when given the calories by 400 calories?

→ Answer :

Data : m = 100 gram

∆= = 8 ℃

Q= 400 calories

Problem : c

a). Q = m. c. ∆=

c = B

9.∆b = -!!

%!!.&

= 0,5 kal/ gram ℃

* So the heat kind of substance was 0.5 cal / g C

4) An object with a heat capacity of 500 joules / oC. What is the heat required to raise the temperature of the object is at 20 oC?

→ Answer :

Data : C = 500 joule

∆t = 20 ℃

Problem : Q

a). Q = C. ∆=

= 500. 20 = 10.000 joule

* So the energy needed by the object of 10.000 joules.

Page 16: Soal Fisika Dan Jawaban

5) For example, heat capacity of a calorimeter is 100 calories. What changes when the calorimeter temperature is 840 joules of heat absorbed?

→ Answer :

Data : C = 100 calories /℃

Q = 840 joule

Problem : ∆=

a). ( 1 calories = 4,2 joule )

Q = 840 joule = &-!-,$ = 200 calories

C = B∆b

∆= = B' =

$!!%!! = 2 ℃

* So the change in calorimeter temperature 2 ℃

6) Water temperature of 500 grams of 10℃ mixed with 200 grams of acid at a temperature 50℃ and then stirring until the temperature of the mixture in a state of equilibrium. Determine the temperature of the mixture in equilibrium, if the heat of water = 1 kcal / kg. ℃ and the heat kind of acid = 0.5 kcal / kg. ℃

→ Answer :

Data : mair = 500 gram

mza = 200 gram

tair = 10℃

tza = 50 ℃

cair = 1kkal/ kg.℃

cza = 0,5 kkal/ kg. ℃

Problem : tx

a). Qlepas = Qserap

mza.cza (tza – tx) = 500 .1 (tx – 10)

5000 – 100 x = 500 tx – 5000

Page 17: Soal Fisika Dan Jawaban

10000 = 600 tx

tx = 16,67℃

* So the temperature of the mixture in equilibrium 16.67 ℃

7). A hollow ball ismade of bronze (c = 18 x de"f/ ℃. At temperature 0℃ its radius is 1 m. If the ball is warmed up to 80℃, calculate the area incremen of the ball’s surface !

Answer :

a) ∆6 = A0 g ∆+

= (4.. \$) (2h) ∆+

= (4.. 12) (2. 18 x 10"D) (80-0) 2$

= 11520 x 10"D . 2$

= 1,15 x 10"$ . 2$

• Thus, the area increment of the ball is 1,15 x 10"$ . 2$

8). An object has temperature of 25℃. Express the temperature of the object in the reamur, fahrenhait, and Kelvin scale !

Answer :

a) In reamur scale (°\) +i 5 4j → +_ = 4 5j +i

= 4 5j (25)°\

b) In fahrenhait scale (℉) +i 5 9j (+l – 32) → +l = 9/5 +i + 32

= 9 5j (2) + 32 ℉

= 77℉

c). In kelvin scale (K)

+i = +m - 273 → +m = +i + 273

= 25 + 273

= 298 K

Thus, 25℃= 20°\ = 77℉ = 298 K

( AMALIA HAJAR D.E )( AMALIA HAJAR D.E )( AMALIA HAJAR D.E )( AMALIA HAJAR D.E )

Page 18: Soal Fisika Dan Jawaban

KD 4.1

1). An object has temperature of 25˚C. Express the temperature of the object in the reamur, fahrenheit, and kelvin scales !

Answer :

• In reamur scale (R˚)

Tc = 5/4 Tr => Tr = 4/5 Tc

= 4/5 (25)˚R

= 20˚R

• In fahrenheit scale (˚F) Tc = 5/9 (Tf – 32) => Tf = 9/5 Tc + 32 = 9/5 (2) + 32˚F = 77˚F

• In kelvin scale (K) Tc = Tk -273 => Tk =Tc + 273 = 25 + 273 = 298 K Thus 25˚C = 20˚R = 77˚F = 298 K

2). A steel wire has length of 100 cm at temperature 30˚C. If the length of the steel now is 100. 1 cm and a = 10ˉ5/˚C, determine the temperature of steel now !

Answer :

λ = 100,1 cm

λ0 = 100 cm

α = 10̄ 5/˚C

λ = λ0 (1 + α ∆T)

100,1 = 100 (1+10̄5(∆T))

100,1 = 100 + 10-3 ∆T

0,1 = 10-3 ∆T

∆T = 0,1 / 10-3˚C

= 100˚C

∆T = T – T0

Page 19: Soal Fisika Dan Jawaban

T0 = 30˚C

100˚C = T-30˚C

T = 130˚C

3). A 4 liter vesel, 95% og its volume is filled with alcohol. The intial temperature is 0˚C then warmed up to 70˚. Determine the alcohol’s volume which split out if the coefficient of length expansion of vessel 0,000011˚C-1 and coefficient of alcohol expansion is 0,001˚C-1 !

Answer :

∆V vessel = o vessel V0 vessel ∆T

= 3 α vessel V0

vessel ∆T

= 3 (0,000011) (4000cm3) (70)

= 9,24cm3

∆Valcohol = oalcohol V0 alcohol ∆T

= oalcohol . 95% V0 vessel

∆T

= (0,01) (95% . 4000) (70)

= 266 cm3

Therefore

Vvessel = V0

vessel + ∆Vvessel

= 4000 + 9,24

= 4009,24 cm3

Valcohol = V0 alcohol

+ ∆Valcohol

= 3800 + 266

= 4066 cm3

Thus, the alcohol’s volume split out is

∆V = Valcohol - Vvessel

= 4066 – 4009,24

= 56,76 cm3

Page 20: Soal Fisika Dan Jawaban

4). A 3 m3 gas, is increased its temperature from 12˚C into 103˚C at aconstant pressure. Calculate the volume of gas now !

Answer

V0 = 3 m3

o = 1/273 C̊-1

∆T = 103˚C - 12˚C

= 91˚C

Therefore

V = {3 m3) {1 + 1/273 C̊-1 (91˚C)} = 4m3

Thus, the volume of gas now is 4 m3

5). A 500 gram body absorbs heat 400 calori so that its temperature increases 4˚C. Determine the spesific heat of that body !

Answer :

Q = mc ∆T

C = Q m ∆T

because Q =400 calorie, m = 500g , and ∆T = 4˚C, then

c = -!!

� !!��-� kal/g˚C

= 0,2 kal/ g˚C

Thus, the spesific heat of body is = 0,2 kal/ g˚C

6). as erect as berpenampang's armor little along 20 meter has temperatures 20C be heated until 40C. then is iced until 30. get do length difference erect that to long early?

Answer :

a) after been heated armor length increases as big as

∆l = l0 a ∆T

= (20) (12 x 10-6) (40-20) = 4,8 x 10-3 m = 4,8 mm

b) after been iced will happen length cut back as big as

∆l = l0 a ∆T

Page 21: Soal Fisika Dan Jawaban

= (20) (12 x 10-6) (-30 -20)

= -12 x 10-3 m

∆l = -12mm

so elongated cut back happening as much 12mm

( ANIS FITRIYAH )

Page 22: Soal Fisika Dan Jawaban

1) Helium gas in a room measuring 40m2. Temperature of 270C. with the pressure of 9 Pa. how much the current temperature if the volume of gas into the pressure half-240m3 and initial pressure.

2) brass with an area of 50m2 which will be heated to a temperature of 750c size doubled. If the coefficient of expansion in length 19x10-6 / c. specify the required final temperature.

3) a 5m long aluminum with akn heated from 250C to 2500C temperature. if the length expansion coefficient 0.000023/0c. specify: a. added long-term and final.

Answer :

1) ∆l : l0xŁx∆t

: 5x0,000023x225

: 0,025875

L : l0x∆l

: 5 + 0,025875

: 5,025875

2) P1xV1/T1 x P2xV2/T2

9x40/300 x 4,5x240/T2

T2x1 : 6x300

T2 : 1800.

3) ∆A : A0x x∆t

A-A0 : A0x x(T-T0)

(100-50) : 50x38x10-6(T-75)

50/50x38x10-6 = T-75

0,026x106 : T

75t0,026x106:T

75t26000 : T

T:260750c.

( ANNISA MAHARINDA )

Page 23: Soal Fisika Dan Jawaban

1.The following figure shows the graph between temperature and time of the chunk of ice is heated, f the heat kind of ice = 2100 J / kg º C, the heat

melting of ice = 340 000 J / kg, heat of water 4200 J / kg º C and its mass is 300 grams, then the amount

of heat needed to melt ice into water at the point of dissolution (process B - C) is ....

A. 3.150 joule B. 25.200 joule C. 102.000 joule

D. 130.350 joule The answer: BC Process: Changes in the form of (fused)

kg=

gram=m

0,3

300

kgJ=L /340000

joule=

kgJkg=

Lm=Q

102000

/3400000,3 ×

2. Note the picture on the side!

at a temperature measurement process the measurement results obtained at 40oC. how many greater if thermometer temperature scale is replaced with Fahreinheit?

•32OF •40OF •72OF •104OF

The answer: On a scale of comparison:

(F–32) : C = 9 / 5 F = 9/5 C +32 F =9/5(40)+ 32 F =104

Page 24: Soal Fisika Dan Jawaban

3. on a thermometer x, freezing point of water is 50 degrees y and boiling point of water is 200 degrees y. A. . when an object is measured with a thermometer Celsius, the temperature 40 degrees Celsius, what is that temperature when measured with a thermometer X..............

The answer:

on the Celsius scale: BA = 100 degrees Celsius, 0 degrees Celsius = 100 degrees Celsius on the scale X: BA = 200 degrees X-50 DEGREES X\ = 150 degrees X

40 : ^X = 100: 150 :100 ^X=40 (150)

^X = 40 (150) / 100 =60

so, the PA on the scale of X is 60 gerajat X

TP = TA + PA = 50 + 60 = 110 DERAJAT X

4. ethyl alcohol solution temperature measurements showed 78 degrees celcius.menyatakan this temperature in.................. The answer:

t= 78 0 C

T = T + 273

78 +273

T = 351 K

5. on one day, Fahrenheit thermometer in place in a room shows the number 122 degrees fahrenheit.berapakah number that will show by a thermometer in Celsius ? The answer:

(tf - 32) : tc = 9: 5: ( 122-320): tc = 9:5

90:tc = 9: 5 :9tc = 90 (5)

Tc = 90(5)/9=50 derajat celcius

( ASRI LINDUNG SARI )

Page 25: Soal Fisika Dan Jawaban

KD 4.1

1. An object has temperature of 70°R. express the temperature of the object in the celcius, fahrenheit, and kelvin scales!

Dik : 70°R

Dit : 70°R = … °C

70°R = … °F

70°R = … °K

Answer : 70°R = - x 70° = 87,5°C

70°R = `- x (70°+32°) = 229,5°F

70°R = - x (70°+273°) = 428,75°K

2. A made object from steel have length 2000 cm. How much is long accretion of that steel, if happened change of

temperature equal to 100°C? Dik : l1 = 2000 cm

∆T = 100 °C

α= 12 × 10-6 °C-1

Dit : ∆l = ...?

Answer : ∆l =l1 α x ∆T = 2000 × 12 × 10-6 × 100 = 24 x 10-1 cm 3. At temperature 40° C a broadness iron plate 20 m2. If its temperature is boosted up to become 100° And iron

coefficient of linear expansion C equal to 0,000012/° C, hence determining wide of the iron plate! Dik : A1 = 20 m2

T1 = 40° C T2 = 100° C

∆T = T2 – T1 = 100 – 40 = 60° C

α= 0,000012/° C

β= 2α= 2 × 0,000012/° C = 0,000024 Dit : A2 = ... ?

Answer : A2 = A1 (1 + β x ∆T) = 20(1 + 0,000024 × 60) = 20(1 + 0,00144) = 20 × 1,00144 = 20,0288 m2

4. A canister have volume 1 litre at temperature 30° C. If coefficient long muai of canister 2 × 10-5 /° C, hence determining canister volume at temperature 80° C !

Dik : γ= 3α= 6 × 10-5 /°C

∆T = 80°C – 30°C = 50° C V1 = 1 l

Dit : V2 = ...?

Answer :V2 = V1 (1 + γ x ∆T) = 1 (1 + 6 × 10-5 × 50) = 1 + 0,003 = 1,003 liter

5. How big needed to heat boost up temperature of is one iron which is mass 10 kg from 10° C become 90° C, if type heat iron 500 J / K Dik : m = 10 kg

∆T = 90 – 10 = 80° C

Page 26: Soal Fisika Dan Jawaban

c = 500 J/kg Dit : Q = ...?

Answer :Q = m × c × ∆T = 10 × 500 × 80

= 4 x 105 kJ 6. Iron rasher owning mass 3 kg, heated from temperature 20° C till 120° C. If iron absorb heat equal to 135 kJ.

Determining iron kalor capacities and iron type heat ! Dik : m = 3 kg

∆T = 120° – 20° = 100° C Q = 135 kJ

Dit : a. C = ...? b. c = ...?

Answer : a. Kapasitas kalor besi

C = B∆Y =

%# .!!!%!!°' = 1350 J/°C

b. Kalor jenis besi

c = '9 =

%# !# HI = 450 J/kg °C

7. How many needed to kalor alter 2 ices gram at temperature 0° C become aqueous vapour at temperature 100° C? ( liquid = 4.200 J / kg ° C, KL = 336 J / g, and ME = 2.260 J / g) Dik : m = 2 g = 2 × 10-3 kg

∆T = 100° – 0° = 100° C Ku = 2.260 J/g KL = 336 J/g cair = 4.200 J/kg °C

Dit : Qtot = ...? Answer : Q1 Proses Lebur Q1 = m KL = 2 × 336

= 672 J

Q2 Proses menaikkan suhu

Q2 = m cair ∆T = 2 × 10-3 × 4.200 × 100 = 840 J Q3 Proses penguapan Q1 = m Ku = 2 × 2.260 = 4.420 J

Qtotal = Q1 + Q2 + Q3

= 672 + 840 + 4.420 = 6.032 J

Page 27: Soal Fisika Dan Jawaban

8. Water counted 0,5 kg which is have temperature to 100° C in infusing canister from aluminium owning mass 0,5 kg. If temperature early canister equal to 25° C, aluminium type heat 900 J / kg ° C, and heat type irrigate 4.200 J / kg ° C, hence determining balance temperature which reached! ( assume there no heat emptying into environment ) Dik : mbjn = 0,5 kg

mair = 0,5 kg Tair = 100° C Tbjn = 25° C cair = 4.200 J/kg °C cbjn = 900 J/kg °C

Dit : Ttermal = ...? Answer :

Qlepas = Qterima

m × cair × ∆T air = m × cbjn × ∆T bjn 0,5 × 4.200 × (100 – Ttermal) = 0,5 × 900 × (Ttermal – 25)

210.000 – 2.100 Ttermal = 450 Ttermal – 11.250 2.550 Ttermal = 222.250

Ttermal = $$$.$ !$ !

= 87,156° C

8. known by external and inner surface temperature a window pane owning Length 2 m and wide 1,5 m partake to partake 27° C and 26° C. If is thick of the glass 3,2 thermal conductivity and mm glass equal to 0,8 W / m ° C, determining is fast of late heat stream the window! Dik : d = 3,2 mm = 3,2 × 10-3 m2

A = 2 ×1,5 = 3 m2

∆t = 27 – 26 = 1° C k = 0,8 W/m °C

Dit : H = ...? Answer :

H = k × A ×∆Yp

= 0,8 × 3 × %

#,$ m %!"#

= 750 J/s 9. Air in a room chamber show scale 25° C, while temperature surface of the room chamber glass window 15° C. If

coefficient convection 7,5 × 10-5 Wm-2 (° C)-4, hence determining is fast of absorbed heat by glass window for the width of 0,6 m !

Dik : ∆T = 25 – 15 = 10° C A = 0,6 m² h = 7,5 × 10-5 Wm-2(°C)-4

Dit : H = ....? Answer :

H = h × A × ∆T4 = 7,5 × 10-5 × 0,6 × 104

= 0,45 W

10. A flimsy plate have total wide of surface 0,02 m2. The platein heating with a stove till its temperature reach 1.000 K. If plate emisitas 0,6, hence determining is fast of transmitted radiasithe plate! Dik : A = 0,02 m2

T = 1.000 K e = 0,6

σ = 5,6705119 × 10-8 W/mK4 Dit : H = ...? Answer :

H = Aeσ T4

Page 28: Soal Fisika Dan Jawaban

= 0,02 × 0,6 × (5,6705119 × 10-8) × (1.000)4 = 6.804 W

11. ingot have Mass to 2 kg have temperature 25°C. To boost up its temperature become 75°C required heat equal to 5.104 kal. If its temperature is boosted up to become 125°C hence how much is required heat? Dik : = 2 kg = 2000 gr

∆t1 = 75 − 25 = 50°C Q1 = 5.104 kal

∆t2 = 125 − 25 = 100°C Q2 = ? Type object of heat can be determined from first situation.

Q1 = m c ∆ t1 5.104 = 2000 . c . 50

c = 5 kal/gr°C

∆t2 is : Q2 = m c ∆ t2 = 2000 . 5 . 100 = 105 kal

12. Above saucer there are 100 ices gr have temperature to 0°C. heat molten of ice known equal to 80 kal/gr. If at ice the given by kalor equal to 6000 kal hence how much ices which have melt? Dik : m0 = 100gr

L = 80 kal/gr Q = 6000 kal

molten Ice mass can be determined as follows. Q = m L 6000 = m . 80 m = 75 gr

molten Ice mass is 75 gr mean the percentage of equal to = × 100 % = 75 %

13. 20 ices gr have temperature - 5°C and pressure 1 atm given by heat till become water have temperature to 80°C.

Type heat irrigate 1 kal / gr°C, ices type kalor 0,5 kal / molten heat and gr°C of ice 80kal / gr. How much is passed to heat is the ice?

Q = Q1 + Q2 + Q3

= ms cs Δts + m L + ma ca Δta

= 10 . 0,5 . (50) + 20 . 80 + 20 . 1 . (80O) = 50 + 1600 + 1600 = 3250 kal

14. Thermos bottle contain 230 gram coffee at temperature 80°C.Is then enhanced by milk counted 20 gram have

temperature 5°C. Otherwise there is mixing heat and also permeated by heat is thermos bottle and coffee type kalor

= milk = water = 1,00 kal / °C g, hence how much is temperature mixture balance? tK = 80 OC, mK = 250 gr tS = 5 OC, mS = 20 gr

c = 1 kal/gr °C

QS = QK

mS cS ΔtS = mK cK ΔtK

20 . 1 . (t - 5) = 230 . 1 (80 - t) 250 t = 18400 + 100 t = 74°C

15. In glass contain 200 cc irrigate 40°C is then included 40 ices gram 0°C. If glass heat capacities 20 kal/ molten And heat ice is 80 kal / gr, hence how much is temperature as well-balanced as?

ma = 200 gr, ta = 40°C Cg = 20 kal/C, tg = ta ms = 40 gr, ts = 0°C

Page 29: Soal Fisika Dan Jawaban

Ls = 80 kal/gr

Q1 + Q2 = Q3 + Q4

ms Ls + ms ca Δts = Cg Δta + ma ca Δta

40 . 80 + 40 . 1 . (t - 0) = 20(40 - t) + 200. 1 . (40-t) 260 t = 8800 - 3200

t = 21,6°C

( AZIZAH ARDYANTI )

Page 30: Soal Fisika Dan Jawaban

1.hot at 12 kj is given on the piece of metal that has a mass of 2500 grams 30oC. If the heat kind of metal is 0.2 calories / groC, determine the final temperature of the metal! Discussion :

D1: Q = 12 kilojoule = 12000 joule m = 2500 gram = 2,5 kg

T1 = 30oC

c = 0,2 kal/groC = 0,2 x 4200 joule/kg oC = 840 joule/kg oC

D2 : T2 =...?

D3 : Q = mc∆T 12000 = (2,5)(840)∆T

∆T = 12000/2100 = 5,71 oC

T2 = T1 + ∆T = 30 + 5,71 = 35,71 oC

2. 500 grams of ice temperature of-12oC heated to a temperature-2oC. If the heat kind of ice is 0.5 cal / GOC, specify a lot of heat is needed, stated in joules! Discussion :

D1 : m = 500 gram

T1 = −12oC

T2 = −2oC

∆T = T2 − T1 = −2o − (−12 ) = 10oC

c = 0,5 kalori/groC D2 : Q = ....? D3 : Q = mc∆T Q = (500)(0,5)(10) = 2500 kalori 1 kalori = 4,2 joule Q = 2500 x 4,2 = 10500 joule 3. 500 grams of ice-temperature 0oC would be disbursed until the overall temperature water ice 0oC. If the heat kind of ice is 0.5 cal / GOC, and the heat melting ice is 80 cal / g, determine a lot of heat is required, state in kilocalories! Discussion :

D1: m = 500 gram L = 80 kalori/gr

Page 31: Soal Fisika Dan Jawaban

D2 : Q = ....? D3 : Q = mL Q = (500)(80) = 40000 kalori = 40 kkal

4. 500 grams of ice-temperature liquefied 0oC going to be 5oC temperature water. If the heat kind of ice is 0.5 cal / GOC, heat melting ice is 80 cal / g, and heat type of water 1 cal / GOC, specify a lot of heat is needed! Discussion :

D1: m = 500 grams liquid = 1 calorie / groC Les = 80 calories / gr The final temperature → 5oC D2: Q = .....? D3: To make ice to become water 5oC 0oC there are two processes that must be passed: → The process of melt ice into water 0oC 0oC temperature, heat is needed call Q1 Q1 = mLes = (500) (80) = 40000 calories → The process of raising water temperature to be 0oC 5oC water, heat required call Q2 Q2 = mcair∆Tair = (500) (1) (5) = 2500 calories Total Heat required: Q = Q1 + Q2 = 40000 + 2500 = 42500 calories

5.500 grams of ice-temperature-10 ° C would be disbursed until 5oC temperature water. If the heat kind of ice is 0.5 cal / GOC, heat melting ice is 80 cal / g, and heat type of water 1 cal / GOC, specify a lot of heat is needed! Discussion :

D1: m = 500 grams ces = 0.5 calories / groC liquid = 1 cal / groC Les = 80 cal / gr The final temperature → 5oC D2: Q = .....? D3: To make the ice - 10 ° C until the water 5oC there are three processes that must be passed: → The process to raise the temperature of the ice from the ice-temperature-10 ° C 0oC, heat is needed call Q1 Q1 = mces∆Tes = (500) (0.5) (10) = 2500 calories → The process of melt ice into water 0oC 0oC temperature, heat is needed call Q2 Q2 = mLes = (500) (80) = 40000 calories → The process of raising water temperature to be 0oC 5oC water, heat required call Q3

Page 32: Soal Fisika Dan Jawaban

Q3 = mcair∆Tair = (500) (1) (5) = 2500 calories Total Heat required: Q = Q1 + Q2 + Q3 = 2500 + 40000 + 2500 = 45000 calories

6.200 grams 80oC temperature water mixed with 300 grams of water temperature of 20oC. Determine the temperature of the mixture! Discussion :

D1: m1 = 200 gram

m2 = 300 gram

∆T1 = 80 − t

∆T2 = t − 20

D2 : Suhu akhir = t = ......? D3 : Qlepas = Qterima

m1c1∆T1 = m2c2∆T2

(200)(1)(80 − t) = (300)(1)(t − 20) (2)(1)(80 − t) = (3)(1)(t − 20) 160 − 2t = 3t − 60 5t = 220

t = 44oC

( AZIZAH HANDAYANI )

Page 33: Soal Fisika Dan Jawaban

1. The temperature of a body temperature of 80oC declare the object in degrees Reamur and degrees Fahrenheit. Completion:

Unknown : t = 80oC

Asked : a) oR = ...?

b) oF = ...?

answer :

a) C: R = 5: 4 b) C: (F – 32) = 5: 9

80: R = 5: 4 80: (F – 32) = 5: 9

5 R = 320 5(F – 32) = 720

R = 64oR 5F – 160 = 720

Jadi 80oC = 64oR 5F = 880

F = 176

Jadi 80oC = 176oF

2. Celsius thermometer and Reamur used to measure the temperature of a objects turned out to scale the amount indicated by the two thermometers = 90o. How OC and OR temperature object?

Completion :

Unknown : C + R = 90o

Asked : t dalam oC dan oR

Answer :

C + R = 90

R = 90-C

C: R = 5 : 4

C: (90 – C) = 4C

450 – 5C = 4C

450 = 9C

C = 50

R = 90 – C

R = 90 – 50 = 40

Page 34: Soal Fisika Dan Jawaban

Jadi suhu benda tersebut: 50oC dan 40oR

3. How many calories of heat needed to heat 2 liters of water from 30oC to 80oC when the density of water = 1 gram/cm3 and heat of water = 1 cal / groC? Completion:

Unknown : V = 2 liter = 2 . 103 cm3

∆t = 80oC – 30oC = 50oC

ρ = 1 gram/cm3

c = 1 kal/groC

Asked : Q = ...?

Answer : m = ρ . V = 1 x 2 x 103 = 2 . 103 gram

Q = m . c . ∆t

Q = 2 . 103 . 1 . 50

Q = 105 kalori

4. A total of 2 kg of water is heated so that its temperature rises from 15 C to 40 ᵒ ᵒ C. If the heat of water 4200 J / kg o C, how much heat is absorbed by water? Completion:

Unknown : mair = 2 kg

T1 = 15ᵒC

T2 = 40ᵒC

c =4.200 J / kgᵒC

Asked : C =........?

Answer : : Q = m.c.∆T

= (2) (4.200) (40 – 15)

= 210.000 J

5. A first wire length 2 m with an initial temperature 20ᵒC. The wire is heated so that its temperature rose to 70ᵒC. If the wire is made of materials that length expansion coefficient 10-5 ᵒC-1, compute: a. Added length of the wire! b. The length of wire after heating! Completion:

a. ∆ℓ = ℓo h ∆T

= (2) (10-5) (70 – 20)

= 10-3

= 0,001 m

Page 35: Soal Fisika Dan Jawaban

b. ℓ’ = ℓo + qℓ = 2 + 0,001

= 2,001 m

6. A block of mass 2 kg has a heat capacity of 500 J / ᵒ C. If such material is received heat by 2,000 J, compute: a. Beam temperature rise! b. Heat the type of beam! Completion:

Unknown : M = 2 kg,

C = 500 J / ᵒC,

Q = 2.000 J

Asked : a. Q..........?

b. C..........?

Answer : a. Q = C ∆T

∆T = B' =

$.!!! !! = 4ᵒC

b. C = m c

c = '9 =

!!$ = 250 J / kgᵒC

( CLARIN PUSPA )

Page 36: Soal Fisika Dan Jawaban

1. Problem solving

In how many temperature of the Celcius and Farenheit scale show the same value ?

Solution :

Use equation C = 5/9 ( f- 32 ) atau F = 9/5 (C+ 32 )

In this case, C = F so C = 5/9 ( C-32 )

9/5C = C – 32

9 C – 5 C = -32

5

4C = -160

C = -40 C

So, Celcius and Farenheit scale show the same value at temperature -40 C or -40 C

2. Problem solving

A mercury thermometer shows the coloumn length is 2 cm if one of the tips is immerse on the ice that is melting,

and shoes the 12 cm length when it is immersedon thee water that is boiling . if it is immersed on the tea , the

coloumn length is 6 cm. how many the temperature of the tea based on the Celcius scale?

Solution :

The differentiation of the column is 12 cm – 2 cm = 10 cm. the column length is 10 cm showa temperature 100 c. if

the differentiation the column is: 6 cm – 2 cm = 4 cm, the temperature will be

4/10 x 100 C = 40 C

So the temperature of the tea is 40 C

6 CM – 2 CM = 4 cm, suhu yang ditunjukkan 4/10 x 100 C = 40 C

Jadi, suhu air the adalah 40 C

3. Problem solving

The length of the metal stick is 100 cm at temperature 25 C . if metalstick have the cooficient of linear expansion

1.33 x 10 -5/C. determine the length of the stick at temperature 100 C.

Solution :

We have ΔT = 100 C – 25 C = 75 C , L0 = 100 cm and α = 1,33 x 10 -5/ C

Use the equation

Lt = lo (1 + α ΔT )

Lt = 100 ( 1 + 1.33 x 10 x 75 )

= 100 ( 1+ 0.0009 )

= 100.1 cm

Page 37: Soal Fisika Dan Jawaban

4. Problem solving

A vessel has the volume 1 litre at temperature 25 C. if vessel have the coefficient of linear expansion 2 x 10 -5

/C,

determine the volume of the vessel at temperature 75 C.

Solution :

We have γ = 3 α = 6 x10 -5

/C

ΔT = 75 C – 25 C = 50 C

V0 = 1 litre

Use the equation Vt = V0 (1 + γ ΔT )

= 1 ( 1 + 0.00006 x 50 )

= 1 ( 1+ 0,003 )

= 1,003 litre

5. Problem solving

At the temperature 27 C. a gass with volume V. how C does that gas have to be heated in the constant preasure in

order to make the volume be twice ?

Solution :

Vt = V0 (1 + γ ΔT )

2 vo = 1 + ΔT/273

1 = ΔT/273

ΔT = 273 C.

In conclude, gass has to be heated until temperature

T = T0 + ΔT

= 27 = 273

= 300 C

6. Problem solving

Convert 55 F to the Celcius and the Kelvin temperature scale

Solution :

First convert from F to C

Tc = 5/9 (F- 32 )

= 5/9 ( 55 - 32)

= 13 C

Next, convert F to K you can convert from C to K

Tk = TC + 273

Page 38: Soal Fisika Dan Jawaban

= 13 + 173

= 286 K

7. Problem solving

The Eiffel Tower, connstruted in 1889 by Alexander Eiffel, is an impressive latticework structure made of iron. If the

tower is 301 m high on a 22 C day, how much does is height decrease when the temperature cools to 0,0 C? (

coefficient of linear expansision for iron is 12 x 10-6/ K)

Solution :

We have Lo = 301 m; To = 22C; T1 = 0,0 C; αiron = 12 x 10-6/ K

ΔL = α LoΔT

= (12 x 10-6/ K)( 301)(0.0C – 22 C )

= -0,079 m

= - 7,9 cm

8. Problem solving

The heat capacity of 1.00 kg of water is 4,186 J/K. what is the temperature change of the water if 505 J of heat is

added the system

Solution :

We have m = 1.000 kg

C = 4,186 J/K

Calculate ΔT fo Q = 505 J

ΔT = Q/C

= 505/4.186

= 0,121 K

9. Problem solving

The heat capacity of 1.00 kg of water is 4,186 J/K. what is the temperature change of the water if 1,010 J of heat is

removed

Solution :

Since heat is removed in this case Q = -1,010 J

ΔT = Q/C

= -1,010/ 4,186

= 0,141 K

Page 39: Soal Fisika Dan Jawaban

10. Problem solving

One of the window in a house has the shape of a square 1.0 m on a side. The glass in the window is 0.50 cm thick.

How much heat is lost through this window in one day if the temperature in the house is 21 0C and the temperature

otside is 0.0 oC ? ( thermal conductivity of glass is 0.84 W/m.K )

Solution :

We have L = O.50 cm = 0.005 m

T1= 210C

T2= 0.00C

K= 0.84 W/m.K

T= one day = 86.400 s

Recall that 1 W = 1 J/s

The square of the window

A = S2

= ( 1.0 )2

= 1.0 m

2

The heat is lost this window in one day is

Q = kA r∆YC s = = 56 ra%"a$

C s =

= (0.84 )( 1 )r$%"!!.!! s =

= 3.0 x 10.8 J

11. Problem solving

Water 1 litre with the temperature 20 0C is heated until boling. Calculate the heat that is needed

Solution :

We have

The Water mass (m) = 1 kg ( 1 litre water mass in the temperature 20oC = 1 kg )

The Water temperature at frist (T0) = 20 0C

The amount of the heat is

Q = c m ΔT

Q = 1 x 1 x 80 = 80 kcal

Q = 80,000 x 4.2 J = 336 Kj

Page 40: Soal Fisika Dan Jawaban

12. Problem solving :

A stick of iron with 1 kg mass has the temperature 20 0C. the iron temperature is creasing until 30

0C with an electric

heater with power 1 kW. If the heater efficien is 100 % the time that is needed 20 second, calculate the iron specific

heat

Solution :

We have

ΔT = 30-20=10 0C

M =1 kg

P = 1 kW

Q = c m ΔT

Q = 1 Kw x 20 sekonds, remember , joule = watt seconds

Q= 20.000 joule. Add this value in the specific heat equation

C = Q/ ΔT

= 20.000/ 1 x 10 = 2000 J/kg 0C

( DEVI RATNA SAFITRI )

Page 41: Soal Fisika Dan Jawaban

• 1. The temperature of an object expressed in the Celsius scale is 25oC. State theobject temperature in Fahrenheit scale! Answer:

• at – #$ $%$"#$ =

$ "! %!!"!

• at – #$ $%$"#$ =

$ %!!

TF = v $ %!! � 180w + 32

= 77 oC 2. An initial wire length 2 m with an initial temperature 20oC. Wire in the heat so thetemperature rose to 70oC. If the wire is made of materials that length expansioncoefficient 10-5 oC-1, calculate the added length of wire. Answer:

• △ ℓ = ℓ0 . h . ∆+ = (2) (10-5) (70 – 20 ) = 10-3 = 0,001 m

3. A total of 2 kg of water in the heat so the temperature rise of 15oC to 40oC. If theheat of water 4200 J / kgs 0C, how much heat is absorbed by water? Answer:Mair = 2 kg

T1 = 150C T2 = 40oC C = 4200 J / kg 0C

• Q = m . c . ∆+ = (2) (4200) (40-15) = 210.000 J

• ay – $)# #)#"$)# =

#!"! %!!"!

• ay – $)#

%!! = #! %!!

Tk = �30 + 273� = 303 oC

( DIMAS FASIHATIN )

Page 42: Soal Fisika Dan Jawaban

4.1

1)Mercury is heated until boiling, then at evaporates. If kalor yawns mercury 854000 j. / kg and mercury amount that yawns as much 0,8 kg. Determinative foots up kalor that needed to yawn that mercury!

mraksa : 0,8 kg

C : 0,854 x 10-6 J/kg

Answer

Q = m . C

= (0,8)(0,854 x 106)

= 0,683 x 106 J

= 683 kJ

2) One steel plate square have flank 30 cm and 20°C's temperatures. If steel plate is heated until 130°C. Get what extent final steel-plated?

Ao = o2

302 = 900 cm2

T = (130 – 20) = 110°C

hbaja = 12 x 10-6/°C Answer : A = Ao(1 + β T)

= 900(1 + 24 x 10-6.110) = 902,38 cm2 (Far ranging final)

3 ) To raise temperatures 0,5 kg irons as big as 30°C needed by kalor as big as 6750 Joules. Determinative

a. kalor's capacity iron

b. Kalor is iron Type, and

Kalor that needed to raise temperature as big as 40°C

m = 0,5 kg

t1 = 30°C

t2 = 40°C

Q = 6750 kg

Answer :

Page 43: Soal Fisika Dan Jawaban

a. � = B∆Y

= D) ! G#!°8

= 225 J/kg°C

b. � = '9

= $$ !,

= 450 J/kg°C

c. Q = mC∆t

= (0,5)(450)(40)

= 9000 J

4)As erect as berpenampang's armor little along 20 meter has 20°C's temperatures is heated until 40°C, then is iced until 30°C. Get do length difference erect that to long early?

= 20 m

t1 = 20°C t2 = 40°C t3 = -30°C

Answer :

a) After at heats, long armor increases as big as

∆ = 0αqT

=(20)(12 x 10-6)(40 - 20)

= 4,8 x 10-3 m

= 4,8 mm

b) After been iced elongated cut back happening as big as ∆ = 0αqT

=(20)(12 x 10-6)(-30-20) = -12 x 10-3 m

5 ) Waters as much 2 liters and gets 25°C's temperatures is heated in pan until its temperature increases as 50°C. Get do afters water volume at heats? V0 = 2 ℓ γ = 210 x 10-6 ∆T = (273 + 50) – (273 + 25) = 25°K

Page 44: Soal Fisika Dan Jawaban

Answer : V = V0(1 + γ ∆T) = 2(1 + 210 x 10-6. 25) = 2,0105 ℓ

6) One armor cube have flank 1 cm. Account volume that cube on 80°C's temperature if temperature initially 20°C.

0 = 1 cm

,0 = 1 cm3 h = 12 x 10-6/°K γ = 3 x 12 x 10-6 = 36 x 10-6/°C Answer: ∆V = V0 γ ∆T

=(1)(36 x 10-6)(80-20) =2,2 x 10-3

V = ∆V + V0

=0,0022 + 1

=1,0022 cm3

4.2

1) One surface serves a ball berpenampang's iron meempunyai little temperature makes a abode 500°C, other surface temperature diijaga makes a abode 150°C. If log length 10 cm, Get what runaway speed kalor / Satuan's transfer extends to pass that iron log (thermal conductivity 73 w / mK) L = 10 cm = 0,1 m

K = 73 w / mK

∆T =350°C Answer : >= = 5 6 �+2 − +1�

>= = 73 �350�

0,1

= 25550 z/22 = 25,55 5z/22

2) One hot source object is looked on as black body perfect broadly surface 100 mm 2 and 927°C's temperature. Account kalor that was emitted by that object. A = 100 mm 2

T = 927°C = 1200°K

Page 45: Soal Fisika Dan Jawaban

ε = 1 (black body)

δ = 5,67 x 10 -8 W / m 2 °K 4

Answer : Bb = |6}+1

4

5,67 X 10-8 X 0,0001 X 1 X (1200)4

11,75 W = 11,75 J/s

3) One room with spatial coolant (AC) having window-glass that its extent 2,0m x 1,5m and its thick 3,2 mm. If temperature on surface deep 25°C's glass and temperature on surface outer 30°C's glass, get what runaway speed kalor's conduction ingoing goes to that room?

A = (2,0m) (1,5m) = 3,0 m 2

d. = 3,2 mm = 3,2 x 10 -3 m

k = 0,8 w / m°K ∆T = 30°C - 25°�=5°C

Answer :

Bb = H~�Y

p

= �!,&��#,!�� ��#,$ m %!J��

= 3750 W

4)Extensive perfect black body its surface 0,5 m2 and its temperature 27 oC. If its peripheral temperature 77°C, account:

a. kalor that permeated unified extensive unified time

b. totaled energy that is emitted up to 1 hour.

Answer:

Black body, therefore e = 1

T1 = 300 K

T2 = 350 K

δ = 5,672.10-8 watt/m2K4

a. R = e δ( T2 - T1)

= 1. 5,672.10-8 (3504 - 3004)

= 391,72 watt/m2

b. R = B

∆Y

Page 46: Soal Fisika Dan Jawaban

Q = R. A. t

Q = 391,72. 0,5. 3600

= 705060 Joule

5)A stem of 150 cm metal with section area of 30 cm2. One of it’s end touches the ice (0˚C) and another end touches and object with temperature of 100˚C. If it’s thermal conductivity is 0,9 cal/s cm˚C, calculate the heat that propragate passing through the metal in 10 seconds!

K = 0,9 cal/s cm˚C

A = 30 cm2

∆T = 100˚C - 0˚C = 100˚C

d = 150 cm

t = 10 sekon

Answer :

=

= 180 calor

6)A wall has the relative constant temperature 0f 25˚C in outdoor air temperature of 18˚C. Calculate the dissipation heat in 3 hours because of heat convection if the wall area 15 m2 and h = 3,5 J/s m2˚K

h = 3,5 J/s m2

K, A = 15 m2

∆T = 25˚C - 18˚C = 7˚C = 7˚K

t = 3 hours = 10800 s

Answer :

Q = h A ∆T t

= (3,5)(15)(7)(10800)

= 39,69 x 105 J

7)A copper ball of 20 cm2 is warmed up until 127˚C. If its emissivity is 0,4 and � = 5,67 x 10-8 W/m2 K4, calculate it’s energy of radiation per second.

� = 5,67 x 10-8 W/m2 K4

A = 20 cm2 = 2 x 10-3 m2

T = 273 + 127 = 400˚K

Page 47: Soal Fisika Dan Jawaban

e = 0,4

Answer :

= (5,67 x 10-8)(2 x 10-3)(400)4

= 1,16 Watt

8) a fluid is flowing with a flow a longitudinal diameter width of40cm2 . the fluid is flowing from a place with a temperature of 100˚C to 30˚C. If the fluid’s ceefficient of thermal covection is 0,05 cal/m2s C˚. Count the amount of heat propagating per second !

Answer :

H = h A ∆T

= 0,05 x 40 x 10-4 x (100 – 30)

= 140 x 10-4

KD4.3 1)Extensive perfect black body its surface 0,5 m2 and its temperature 27 oC. If temperature its peripheral 77 oC, account:

a. kalor that permeated unified extensive unified time b. totaled energy that is emitted up to 1 hour.

Answer: Black body, therefore e = 1 T1 = 300 K T2 = 350 K = 5,672.10-8 watt/m2K4

a. R = e (T2

4 – T14)

= 1. 5,672.10-8 (3504 - 3004) = 391,72 watt/m2

b. R = B

~ .b Q = R. A. t

Q = 391,72. 0,5. 3600

= 705060 Joule

2) Water as heavy as 100 g by 20°C's temperatures dashed with by temperatures 80 g water gets 70°C's temperatures. Determine waters mixed temperature!

m air1 = 100 g

Page 48: Soal Fisika Dan Jawaban

m water water = 80 g

T 1 = 20°C

T 2 = 70°C

Answer:

Accepted = 20°C

Given = 70°C

Mixed temperature = x°C

Kalor who is given = Kalor is accepted (0,08)(c)(70-x) = (0,10)(c)(x-200

5,6 – 0,08x = 0,1x – 2 0,18x = 7,6

� = 7,60,8 = 42,2°�

3)Water as heavy as 150 g gets 80°C's temperatures is inserted into glass gets mass 150 g by 25°C's temperatures. Temperatures decreased determinative that water is acquired temperature balance kalor's glass water 840°C's glass water!

Balance temperature = x Ta = 80°C Tg = 25°C Ma= mg =150g Ca = 4180 J/kg°C Cg = 840 J/kg°C Answer :

Kalor is taken down = kalor is accepted Maca(Ta – x) = mgcg(x – Tg)

O,15.4180.(80-x) = 0,15.840.(x-25) 50160 – 627x = 126x – 3150

53310 = 753x X = 70,8°C

4)A piece cubic aluminium with mass 200 90°C's gram and temperature is dyed into meaty container 1 kg water gets 20°C's temperature. Balance temperature determinative water aluminium. Kalor is water type 4180 j. / Kg°c and kalor is aluminium type 900J / Kg°C.

Kalor takes down Q= mC∆t = 0,2 x 900 x (90 – x) =180(90-x) = 16200 – 180x Kalor accepts Q= mC∆t =1 x 4180 x (x – 20) =4180x – 83600 Kalor takes down = kalor accepts 16200 – 180x = 4180x – 83600

Page 49: Soal Fisika Dan Jawaban

4360x = 99800 X = 22,8°C

5)If 2 kg water gets temperatures 5 0C be dashed with 5 kg water gets temperatures 26 0C, therefore determine both of mixed final temperature substance! Answer: Since both of substance a sort, therefore kalor type of substance second is as, and can mutually negate. Q grasping = q escapes m1.c.∆t1 = m2.c. ∆t2 2.c.(t-5) = 5.c.(26-t) 2.t – 10 = 130-5.t t = 140 : 7 t = 200C 6)If 0,5 kg ice gets temperature? 10 0C is dashed with a number water gets temperature 40 0C,so its mixture temperature is 20 0C, therefore mass determinative of water that is draughted! (ces = 2100 j. / kg.0C, c water = 4200 j. / kg. K, Les = 3,35. 105 j. / kg) Answer: In this process ice will experience three phase form, which is solid (ice), melt together and form moltens. Be water gets temperature 400C just experience one step, which is temperature change Q grasping (ice) = Q escape (water) Q1 + Q2 +Q3 = Q4 163 m1.ces.∆t1 + m1.L + m1.cair.∆t2 = m2.cair.∆t3 0,5x2.100x(0(10)) + 0,5x3,35.105 + 0,5x4.200x(40-20)= m2x4.200(4020) 10.500 + 1,675.105 + 42.000 = 84.000 x m2

2) m2 =2,61kg

7 ) Alloy get mass 150 gr is heated until 500°C, then is inserted into water that its mass 400 gr and 15°C's temperatures. Water lies in calorimeter that its container made from aluminium with mass 200 gr. Balance temperature that is reached is 28°C. Account kalor that alloy sample type

Answer :

Kalor that detached sample = calorimeter water absorbed heat

m2 c2 (T2 – x) = m1 ca (x – T1) + mw cw (x – Tw)

0,1 x c2 (500 – 28) = 0,4 (4.180)(28 – 15) + 0,2 (900) (28 – 15)

47,2 c2 = 21.736 + 2.340

47,2 c2 = 24.076

c2 = 510 J/Kg˚C

so, kalor is that alloy sample type is 510 j. / Kg°C

( ELMA ELVIANA )

Page 50: Soal Fisika Dan Jawaban

KD 4.1 1. If a much of 75 g water with a temperature 0 0C is mixed to as much of 50 g water 1000C,the final

temperature of the mix is….. 2. A thermometer X,if it is used to measure the boiling water temperature pointing to 1200X and in

the ice pointing to -50X .Then for a temperature of 240X, the thermometer 0C points at……. 3. In how many temperature of the Celcius and Fahrenheit scale show the same value ??? 4. A mercury thermometer shows the column lenght is 2 cm if one of the tips is immersed on the ice

that is melting and shows the 12 cm lenght when it is immersed on the water that is boiling. If it is immersed on the tea,the column length is 6 cm. How many the temperature of the the based on the Celcius scale ???

5. The length of the metal stick is 100 cm at temperature 250C. If metal sticks have the coefficient of the linear expansion 1,33 x 10-5/0C,determine the length of the stick at temperature 1000C……

6. A vessel has the volume 1 litre at temperature 250C. If vessel haave the coefficient of linear expansion 2 x 10-5/0C,determine the volume of the vessel at temperature 750C ………

7. A water vessel at temperature 00C has the volume 100 cm3. If that vessel is filled by the mercury, then heated until 500C,how many does the mercury spilled ??? (h glass = 9 x 10-6 /0C and oraksa = 1,8 x10- 4 /0C )…..

8. At the temperature 270C, a gas with volume V. How 0C does that gas have to be heated in the constant pressure in order to make the volume be twice ??

KD 4.2 1. A metal bar has a longitudinal of 10 cm2 and thickness of 1 m.One of its ends is heated that the

temperature difference of the two ends is 500C. If the thermal conductivity is 0,2 cal/msC0,count the amount of heat propagating per second ???

2. A fluid is flowing with a flow longitudinal diameter width of 20 cm2. The fluid is flowing from a place with a temperature 0f 1000C to 600C. If the fluid’s coefficient of thermal convection is 0,01 cal m2sC0,count the amount of heat propagating per second…..

3. Count the energy radiated by an object processing a temperature of 1270C and its surface width is 2m2 for 1 minute if it is considered to be absolutely black……

4. One of the windows in a house has the shape of a square 1,0 m on a side. The glass in the window is 0,50 cm thick. How much heat it lost through this window in one day if the temperature in the house is 210C and the temperature outside is 0,00C ? (thermal conductivity of glass is 0,84 W/mK)

KD 4.1 1. Q lepas = Q terima

m.c.∆T = m.c.∆T 75(tc-0)=50(100-tc)

75tc=5000-50tc

75tc-50tc=5000

125tc=5000

Tc=400C

2. �"m9��

�9�m"�9�� = '%!!

Page 51: Soal Fisika Dan Jawaban

24 − �−5�120 − �−5� = �

100

29125 = �

100

2900=1250C

23,20=C

3. Use equation C =

` (F-32) or F =

` (F+32)

In the case,C = F , so C = ` (F-32)

` � = � − 32

`'" '

= −32

4C = -160 C = - 400C

4. The differentiation of the column is 12 cm-2 cm= 10 cm. The column leght is 10 cm shows temperature 1000C. If the differentiation of the column is 6 cm – 2 cm = 4

cm, the temperature will be -

%! × 100 = 40°�, so the temperature of the tea is 400C.

5. We have : ∆T = 1000C-250C = 750C ; l0=100 cm and α = 1,33 x 10-5/C0

Use the equation =

~ lT = l0 (1 + α.∆T)

= 100 (1 + 1,33 x 10-5/C0.750C)

= 100 (1 + 0,00099)

= 100,1 cm

6. We have : γ = 3α = 6 x 10-5/C0

∆T = 750C-250C = 500C

V0 = 1 litre

Use the equation = VT = V0 (1 + γ∆T)

= 1 (1 + 6 x 10-5/C0. 500C)

= 1 (1 + 0,003)

= 1,003 litre

Page 52: Soal Fisika Dan Jawaban

7. We have : αglass = 9 x 10-6/C0 = 3α glass =2,7 x 10-5/C0

V0glass = 100 cm3

γmercury = 1,8 x 10-4/C0 and ∆T = 500C

Use the equation = VTglass = V0glass (1+γglass∆T)

= 100 (1+2,7 x 10-5/C0.500C)

= 100 (1+1,35 x 10-3)

= 100,135 cm3

= VT mercury = V0glass(1+γmercury. ∆T)

= 100 (1+1,8 x 10-4/C0. 500C)

= 100 (1+9 x 10-3)

= 100,9 cm3

= The mercury that spilled = VTmercury-VTglass

= 100,9 – 100,135

= 0,765 cm3

8. Use formula VT = V0 ((1+γ.∆T)

In the case, VT = 2V0

VT = V0(1+γ ∆T)

2V0 = V0(1+∆Y

$)#)

2 = 1+∆Y

$)#

1 = ∆Y

$)#

∆+ = 273℃

In conclude,gas has to be heated until temperature

T = T0 +∆+

= 27℃ + 273℃

= 300℃

Page 53: Soal Fisika Dan Jawaban

KD 4.2

1. H : 5 ~∆YC

: 0,2%!×%!-4× !

%

: 0,01 x 4,2 = 0,042 J/s 2. H: h.A.∆+

: 0,01 x 20.10-4 x (100-60)

: 8.10-4 x 4,2

: 3,36 x 10-3 J/s

3. We have : T = 127+273=400 K W = e.|T4 = (1)(5,7 x 10-8) (4004) = 1.459,2 watts/m2

The total radiated energy by the object is :

� E = W x A x t � E = 1.459,2 x 2 x 60 � E = 175.104 J

4. We have : s : 1,0 m L : 0,50 cm = 0,0005m T1 = 210C T2 = 0,00C k = 0,84 W/mK t = one day = 86.400s

The square of the window : A = s2

The heat is lost trough the window in one day is:

� Q = 56 r∆YC s . = = 56 rY%"Y$

C s . =

� Q = (0,84) (1)� $%"!!,!!! �86.400

� Q = 3,0 x 108 J

( INGE NINDIANA IRAWAN )

Page 54: Soal Fisika Dan Jawaban

KD 4.1

1. 367°C= …….. °R

• = 4/5 x 367 °R

• = 293,6 °R

2. Sebatang baja (angka muai linier 10-5/ºC) panjangnya 100,0 cm pada suhu 30ºC. Bila panjang batang baja itu sekarang menjadi 100,1 cm, berapakah suhunya sekarang?

Jawab:

Lt = Lo ( 1 + α ∆t) ∆t = (Lt - Lo) / (Lo α)

∆t = (100,1 -100)/(100.10-5) = 100ºC

∆t = takhir - tawal

100 = takhir - 30

takhir = 130ºC

3. Sebuah tabung terbuat dari gelas (α = 10-5/ºC) pada suhu 20ºC mempunyai volume sebesar

250 cm3. Tabung itu berisi penuh dengan eter ( γ = 5.10-3/ºC). Berapakah cm3 eter akan tumpah jika

tabung dipanasi sampai 120ºC?

Jawab:

Gelas:

Vo = 250 cm3

∆t = 120 - 20 = 100ºC

g = 3α = 3.10-5/ºC

Vt = Vo(1 +€γ ∆t) Vt = 250 (1 + 3.10-5.100) = 250,75 cm3

Eter:

Vº = 250 cm3

∆t= 100ºC

g = 5.10-3/ºC

Vt = Vo (1 + γ€∆t) Vt = 250 (1 + 5.10-3.100) = 375 cm3

Jadi volume eter yang tumpah = 375 - 250,75 = 124,25 cm3

1. 367°C= …….. °R

• = 4/5 x 367 °R

Page 55: Soal Fisika Dan Jawaban

• = 293,6 °R

2. One steel ( number of expansion linear 10-5 / º C) length 100,0 cm [at] temperature 30ºC. If/When bar length become militant that now become 100,1 cm, how much is its temperature now?

Lt = Lo ( 1 + α ∆t) ∆t = (Lt - Lo) / (Lo α)

∆t = (100,1 -100)/(100.10-5) = 100ºC

∆t = takhir - tawal

100 = takhir - 30

takhir = 130ºC

3. A made tube of glass = 10-5 / º C) [at] temperature 20ºC having volume equal to 250 cm3. That tube contain full of ether = 5.10-3 / º C). How much is ether cm3 of pour if hot tube until 120ºC?

Glass:

Vo = 250 cm3

∆t = 120 - 20 = 100ºC

g = 3α = 3.10-5/ºC

Vt = Vo(1 +€γ ∆t) Vt = 250 (1 + 3.10-5.100) = 250,75 cm3

Ether:

Vº = 250 cm3

∆t= 100ºC

g = 5.10-3/ºC

Vt = Vo (1 + γ€∆t) Vt = 250 (1 + 5.10-3.100) = 375 cm3

Become ether volume which pour is = 375 - 250,75 = 124,25 cm3

( Muhammad ainul Yaqin )

Page 56: Soal Fisika Dan Jawaban

1. Sebuah Benda mempunyai suhu 550 C, Nyatakan suhu benda tersebut dalam skala

Reamur, Fanrenheit, dan Kelvin!

( An Object has temperature of 550 C, Express the temperature of the object in the

Reamur, Fanrenheit, And Kelvin Scales! )

@ SOLUTION ! @

a. In Reamur Scales ( 0R )

Tc = 5/4 TR TR = 4/5 Tc

= 4/5 (55) oR

= 440 R.

b. In Fanrenheit Scales ( 0F )

Tc = 5/9 ( TF - 32 ) TF = 9/5 Tc + 32

= 9/5 ( 55 ) + 32 oF.

= 1310 F.

c. In Kelvin Scales ( 0K )

Tc = TK - 273 TK = Tc + 273

= 55 + 273

= 3280 K.

2. A steel wire has length of 100 cm at temperature 300 C, If the length of the steel now is

100,1 cm and α = 10 – 5

/0C, Determine the temperature of steel now!

( Seutas kawat baja mempunyai panjang 100 cm pada suhu 300C. jika panjang baja itu

sekarang 100,1 cm dan α = 10 -5/0C, tentukan suhu baja itu sekarang! )

@ Solution @@ Solution @@ Solution @@ Solution @

a. £ = £o ( 1 + α T )

Cause £ = 100,1 cm, £0 = 100 cm, and α = 10 -5/0C, then :

- 100,1 = 100 ( 1 + 10 -5 ( T ) )

- 100,1 = 100 + 10 -3 T.

Page 57: Soal Fisika Dan Jawaban

- 0,1 = 10 -3 T.

- T = 0,1/10 -3 0C

= 100 0C

- T = T – T0

- T0 = 300 C 1000C = T - 300C

T = 1300C

3. Air sebanyak 500 Gram bersuhu 100C dicampur dengan 200 Gram zat asam, bersuhu

500C, kemudian diaduk hingga suhu campuran dalam keadaan kesetimbangan, tentukan

suhu campuran dalam keadaan setimbang, jika kalor jenis air = 1 Kkal/Kg.0C. dan kalor

jenis zat asam = 0,5 Kkal/Kg 0C.

a. Diketahui : - mair = 500 Gram 0,5 Kg.

-mzat = 200 Gram 0,2 Kg.

-tair = 100C

-tzat = 500C

-Cair = 1 Kkal/Kg0C

-Czat = 0,5 Kkal/Kg0C

b. Ditanya :

-tx …………………………. ???

c. Dijawab :

-Qlepas = Qterima

-mzat . . . . Czat . . . . ( tzat - tx ) = mair . . . . Cair . . . . ( tx – tair )

-200 . 0,5 ( 50 - tx ) = 500 . 1 . ( tx – 10 )

-5000 – 100 tx = 500 tx – 5000

10000 = 600 tx

tx = 16,67 0C.

Page 58: Soal Fisika Dan Jawaban

4. Berapakah kalor jenis suatu zat, bila massa benda tersebut sebesar 100

Gram, suhunya akan naik sebesar 800C bila diberikan kalori sebesar 400

Kalori?

@Solution@@Solution@@Solution@@Solution@

a. Diketahui :

- m = 100 Gram

- t = 8 0C.

- Q = 400 Kalori

b. Ditanya :

- c ……………….. !

c. Dijawab :

- Q = m . c . t

- c = Q/m . t = 400/100 . 8

- 0,5 Kal/ Gram 0C.

( MUHAMMAD FIQHI IBAD )

Page 59: Soal Fisika Dan Jawaban

1. An object has temperature 600C. express the temperature of the object in the reamur, fahrenheit and kelvin

scales !

D1 = 600C

D2 = a. R

b. F

c. K

D3 =

a. R = 4 x 60

5

= 48 0R

b. F = 9 x 60 (+32)

5

= 108 + 32 = 140 0F

c. K = 60 + 273

= 3330K

2. If the iron along the 20 m long with a coefficient of expansion 1.2. 10-5 / K temperature is heated from 0

0C to 100 0C, then specify the increment length !

D1 = lo = 20 m

α = 1.2 . 10−5 / Κ

t1 = 0 0C

t2 = 100 0 C

D2 = ∆l

D3 =

∆l = lo . α . ∆t

∆l = 20 . 1,2.10-5. (100– 0)

∆l =2,4.10-3 m

3. Glass area of 2 m2, with a long expansion coefficient 8,5.10-6 K, having heating from 20 ° C to 120 ° C.

Determine the broad end of the glass!

D1 = A0 = 2 m2

Page 60: Soal Fisika Dan Jawaban

β = 8,5.10-6 K

t1 = 200C

t2 = 1200C

D2 = At

D3 =

At =Ao (1+ b.∆t)

A t =2 (1+ 2 x 8,5.10-6.(120 – 20 )

A t =2,66 m2

4. A metal rod made of aluminum length 2 m in 30 ° C. When the long expansion coefficient of aluminum

25 x 10-6 / ° C. What is the added length of the aluminum rod when temperature is raised to 50 ° C!

D1 = lo = 2 m

α = 25 x 10-6 / ° C

t1 = 30 ° C

t2 = 50 ° C

D2 = ∆l

D3 = ∆l = lo . α . ∆t

= 2 . ( 25 x 10 -6 ) . (50 – 30 )

= 10 -3 m

∆l = 0,1 cm

5. A brass block has a length of 5 m, height 2 m, and width of 1 m at 20 ° C. If the heat kind of brass 1.8. 10-

5 / K, determine the volume of brass at a temperature of 120 ° C!

D1 = length = 5 m

Height = 2 m

Width = 1 m

γ = 1,8.10-5

Page 61: Soal Fisika Dan Jawaban

t1 = 20 ° C

t2 = 120 ° C

D2 = Vt

D3 = Vt =Vo (1+ g.∆t) Vt =(5 x 2 x 1) (1+3.x1,8.10-5.(120 – 20 )

Vt =10,054 m3

6. An object has temperature 400R. Express the temperature of the object in the celcius, Fahrenheit, and

Kelvin scales!

D1 = 400R

D2 = a. C

b. F

D3 =

a. 5 x 40

4

= 500C

b. 9 x 40 (+32)

4

= 90 +32 = 122 oF

7. Boiling water temperature 30 ° C thermometer X has the TTA and TTB 150 ° X -50 ° X Y thermometer has the TTA and TTB 130 ° Y 30 ° Y Determine how the water temperature boiling thermometer according to X and Y!

D1 = Boiling water temperature = 30 ° C

TTAX = 130 °X

TTBX = -50 °X

TTAY = 130 °Y

TTBY = 30 °Y

D2 = temperature thermometer X and Y

D3 =

Thermometer X

Page 62: Soal Fisika Dan Jawaban

C – TTBC = X - TTBC

TTAC - TTBC TTAX – TTBX

30 °- 0 = X + 50°

100° 150° +50°

30 ° = X + 50°

100 200

30° = X + 50°

2

X + 50° = 60°

X = 10°

Thermometer Y

30° - 0° = Y - 30°

100° - 0° 130° - 30°

30° = Y - 30°

100 100

Y - 30° = 30

Y = 60 Y

( Muhammad rizqi alfian )

Page 63: Soal Fisika Dan Jawaban

1. Normal human body temperature was 98.6oF. What is the normal human body temperature in Celsius scale? Answer :

98, 6oF = ………. oC

= 5/9 x (98,6 – 32)

= 5/9 x 66,6 = 37 oC

2. The temperature of the boiling point of nitrogen is 77.35oK. State the temperature in Celsius and Fahrenheit scale? Answer :

• 77, 35oK = ………oC = 77,35o – 273o = - 195,65o C = 195, 65oC

• 77, 35oK = ……….oF = 77, 35o + 32o

= 109, 35oC

3. How much heat is required to raise the temperature of 10 kg of water from 20 oC to 80 oC? heat of water = 4180 J / kg oC? Answer : D1 : m = 10 kg T1 = 20 oC T2 = 80 oC Cair = 4180 J/kgoC D2 : Q……??? D3 : Q = m . C . ∆T = 10 . 4180 . 60 = 2508 x 103J

REMIDI KD 4.2

1. Copper triangle has an area of 50 m2 with temperatures up to 10oC with 60 s. If emisitas 0.4 and σ = 5, 67 x 10 W / m2 K, how much radiation energy per second ? Answer : D1 : A = 10m2

T = 10 oC = 283oK e = 0,4 σ = 5, 67 x 10-8 W / m2 K t = 15 s D2 : P….???? D3 :

P = Q / t = e . σ . A . T4

= Q/15 = 0,4 . 5, 67 x 10-8 . 10 . 2834

= 221291,553 watt

Page 64: Soal Fisika Dan Jawaban

2. A stem of 500 m copper with an area 50 m2. One of its and touches the ice 20oC and touches an object with temperature of 50 oC. if its thermal conductivity is 0,9 W/m K. how much head that propagate passing throught the metal in 15 second ??? Answer : D1 : ℓ = 500 m 6 = 50 m2

T1 = 20oC = 293 oK T2 = 50oC = 323 oK K = 0,9 W/m K t = 15 D2 : H….??? D3 : H = Q/t = K . A . (T2 – T1)/ ℓ = Q/15 = 0,9 . 50 . (323-293)/500 = Q/15 = 2,7 = Q = 2,7 . 15 = 40, 5 J So, H = Q/ t = 40, 5/15 = 2,7 J/s

3. What speed the flow of heat through the glass window which an area 2,0 m x 1,5 m and thick 3,2 mm, if the temperature at the surface of the inner and outer windows 15 oC and 30 oC with 10 s and a thermal conductivity of 0.84 J/s m oC ? Answer : D1 : L = 2, 0 x 1, 5 = 3 m A = 3, 2 mm = 0, 0032 m2

T1 = 15 oC T2 = 30 oC K = 0.84 J/s m oC t = 10 s D2 : Q….?? D3 : H = Q/t = K . A . (T2 - T1)/ L = Q/10 = 0, 84 . 0,0032 . (30-15)/3 = Q/10 = 0,01344 = Q = 10 . 0,01344 = 0, 1344 J

4. A thick pieces 3 cm of steel has a section area of 1000 cm2. If one side has a temperature of 100 oC

and the other side has a temperature of 120 oC, with time what is the heat that moves through these pieces every second? With time 20 s and thermal conductivity (k) steel = 40 J / s m Co ? Answer : D1 : L = 3 cm = 0, 03 m A = 1000 cm2 = 0,1 m2

T1 = 100 oC T2 = 120 oC K = 40 J/s m oC t = 20 s D2 : H…….?? D3 : H = Q/t = K . A . (T2 - T1)/ L = Q/20 = 40 . 0,1 . 20/0,03 = Q/20 = 2666,66667 = Q =2666,66667 . 20 = Q = 53333,3334 J

Page 65: Soal Fisika Dan Jawaban

So, H = Q/t = 53333,3334/20 = 2666,66667 J/s

5. A square-shaped object with a length of 4 cm2 side has a temperature of 500 oC and has 50 minutes.

If the object can be considered as a black body, determine the energy radiated per second ...( σ = 5,67 x 10 -8 W/ m2. K4) Answer : D1 : L = 4 cm2 = 0, 0004 m2

T = 500oC = 773 oK t = 10 minutes = 600 s e = 1 σ = σ = 5,67 x 10 -8 W/ m2. K4

D2 : P…..?? D3 : P = Q/t = e . σ . A . T4

= Q/600 = 1 . 5,67 x 10 -8 . 0, 0004 . 7734

= Q/600 = 81,4053257 = Q = 81,4053257 . 600 Q = 48843, 1954 J So, P = Q/t = 48843, 1954/600 = 81,4053257 watt

6. An athlete is to cool down in silence in the locker room with a temperature 15oC. If the athlete's body temperature was 34oC with skin surface area 1.5 m2. Determine the radiation given athlete's body (body emission coefficient = 0.7). Answer : D1 : T1 = 15oC = 288 oK T2 = 34oC = 307 oK A = 1.5 m2

σ = 5,67 x 10 -8 W/ m2. K4 e = 0,7 T2 – T1 = 307 – 288 = 19oK D2 : P….??? D3 : P = Q/t = e . σ . A . T4

= Q/t = 0, 7 . 5,67 x 10 -8 . 1.5 . 194

= Q/t = 0,07799712 J/s

7. A rectangular thin steel plate with side length 10 cm, is heated in a furnace so that its temperature reaches 727oK. Determine heat radiation energy every second. Answer : D1 : A = 2(10 cm)2 = 0,02 m2

e = 1 T = 727 +273 = 1000 oK σ = 5,67 x 10 -8 W/ m2. K4 D2 : W :…? D3 : W = e . σ . A . T4

= 1 . 5,67 x 10 -8 . 0,02 . 10004

= 1134 J

Page 66: Soal Fisika Dan Jawaban

8. Incandescent spiral wire 50 mm2 surface area, temperature 1127oC, 60 % of the electrical energy delivered to the light radiated in the form of heat, and filament are as black objects. How many ampere current flowing in the lamp socket is connected with (voltage 220 V), for the lights to work? Answer : D1 : A= 50 mm2 = 50 x 10-6 m2 T (1127+273) = 1400 oK e = 1 σ = 5,67 x 10 -8 W/ m2. K4 60% daya listrik menjadi radiasi kalor D2 : i. ? D3 :

♦ P = e . σ . A . T4

= 1 x 5,67 x 10-8 x 50 x 10-6 x (1400)4 = 10,89 Watt

♦ P listrik = 10,89/60%= 10,89/0,60 = 18,15 W ♦ P listrik = V.i = 200 x i= 10,15, i = 18,15/220 = 0,0825 A

REMIDI KD 4.3

1. Calculate the amount of head required to melt 150 gram, has a temperature ice 15 oC into water 25

oC, if Cwater = 4200J/kg K, Cice = 2100 J/kg K and Lice = 144000K/kg. Answer : D1 : M = 150 gr = 0,15 kg T 1 = 15o C = 15 + 273 = 288 oK T 2 = 25 oC = 25 + 273 = 298 oK Cwater = 4200J/kg K Cice = 2100 J/kg K Lice = 144000K/kg.

D2 : Q total …?? D3 :

Q ice + Q water + L ice = m.c.∆t + m.c.∆t + L ice = 0,15 . 2100 . 10 + 0,15 . 4200 . 10 + 144000 = 3150 + 6300 + 144000 = 153450 J 2. if 50 grams of water has a temperature of 10oC was mixed with 25 grams of water with a temperature

of 80oC then the final temperature of the mixture is ... D1 : M1 = 50 gr M2 = 25 gr T1 = 10oC T2 = 80oC

D2 = Tc…?? D3 = Q lepas = Q terima m1.c1.∆t1 = m2.c2.∆t2 50.(Tc – 10) = 25.(80 – Tc) 50 Tc – 500 = 2000 – 25 Tc

Page 67: Soal Fisika Dan Jawaban

50 Tc + 25 Tc = 2000 + 500 75 Tc = 2500 Tc = 2500/75 = 33, 333 oC

3. A piece of ice cube of mass 0.2 kg, ice cubes are mixed with hot tea. Mass = 0.2 kg of hot tea. ice temperature = -10 ° C, while the temperature of hot tea = 40 oC. After sticking for some time, ice stone and warm water is mixed into a closed system, what is the temperature after the intervention ? D1

Massa es batu = 0,2 kg Massa teh hangat = 0,2 kg Kalor jenis (c) air = 4180 J/kg Co Kalor jenis (c) es = 2100 J/kg Co Kalor Lebur (LF) air = 334 x 103 J/Kg Suhu es batu (Tes batu) = ‐10 oC Suhu teh hangat (T teh hangat) = 40 oC D2 : Suhu campuran = ? D3 :

Steps first : • Q lepas = mc∆T

= (0,2 kg) (4180 J/Kg Co) (40 oC – 0 oC) = (0,2 kg) (4180 J/Kg Co) (40 oC) = 33.440 Joule = 33,44 kJ

Head accepted by 0,2 kg ice cube for the ascend of its temperature from ? 10 oC until 0 oC. • Q terima = mc∆T

= (0,2 kg) (2100 J/Kg Co) (0 oC – (‐10 oC)) = (0,2 kg) (2100 J/Kg Co) (10 oC) = 4200 Joule = 4,2 kJ

needed to heat melt 0,2 kg ice cube ( needed to heat alter all ice cube become water) • Q lebur = mLF

= (0,2 kg) (334 x 103 J/Kg) = 66,8 x 103 Joule = 66,8 kJ

Pursuant to result of calculation above, obtained by the following result : • Q lepas – Q terima – Q lebur =33,44 kJ - 4,2 kJ- 66,8 kJ

= - 37,56 kJ

So, final temperature below o0C or fixed o0C

4. if 30 grams of water has a temperature of 15oC was mixed with 20 grams of water with a temperature of 50oC then the final temperature of the mixture is ... D1 : M1 = 30 gr M2 = 20 gr T1 = 15oC T2 = 50oC

D2 = Tc…?? D3 = Q lepas = Q terima m1.c1.∆t1 = m2.c2.∆t2

30.(x-15) = 20 (50 – x) 30x – 450 = 1000 – 20x 30x + 20x = 1000 + 450 50x = 1450 X = 1450/50 X = 29oC

Page 68: Soal Fisika Dan Jawaban

5. the ice 50 grams in temperature 10 oC mixed with ice 30 grams in the temperature 50 oC. determine

the mixture temperature of both ice. (C ice = 2100 cal / gr). D1 : M1 = 50 gr M2 = 30 gr T1 = 10oC T2 = 50oC D2 = Tc…?? D3 = Q lepas = Q terima m1.c1.∆t1 = m2.c2.∆t2

50 . (x-10) = 30 (50-x) 50x – 500 = 1500 – 30x 50x + 30x = 1500 + 500 80x = 2000 X = 2000/80 X = 25oC

6. the mass of alumunium 0,5 kg and the temperature 100oC is added to the calorimeter. In the temperature 40 oC, the thermal balance condition. If the water mass in the calorimeter 0,5 kg and the water temperature at first 25oC, determine the alumunium specific heat. (Cair = 1 kkal/kg Co) Answers : D1 : mal = 0,5 kg T1 = 100 oC T2 = 40 oC mwater = 0,5 kg T1 = 25oC D2 : Q…?? D3 : Qlepas = Qterima

mal . Cal . ∆twater = mwater . Cwater . ∆twater 0,5 . Cal .. (100 – 40 ) = 0,5 . 1 . ( 40 – 25 ) 0,5 . Cal .60 = 0,5 . 1 . 15 Cal . 60 = 15 Cal = 15/60 Cal = 0,25 kkal/kg Co

( MUHIMMATUS SYARIFAH )

Page 69: Soal Fisika Dan Jawaban

4.1

1. The temperature of an iron rod with 10 kg mass is going to be raised by 5 K . If the heat needed is 25 kj , how much is the specific heat ? Answer :

� m = 10 kg T = 5 K Q = 25 kj

� (a) =C = Q = 25.000J = 500 J/kg.K M . ∆+ (10kg)(5 K)

2. How much out grow if calor that at needs to thaw ice as much 500 g , on temperature 0oC as water the lot that 0oC ? if at knows calor laten ice fusion becomes water as big as 10 kal / gr ? Answer :

� D1 = m = 500 gr. L = 10 kal/gr

� D2 = Q……..? � D3 = Q = m . L

= 500 .10 =5000 kal

3. One object with calor’s capacity 1000 Joule / oC . How much calor one at need to raise that object temperature as big as 50oC ?

Answer :

� D1 = C = 1000 Joule / oC

=∆+ = 50oC

� D2 = Q ……?

� D3 = Q = c . ∆+

= 1000 . 50

=5.000 Joule

( NANDA PRADANA )

Page 70: Soal Fisika Dan Jawaban

REMIDI KD. 4.1

1. An object has 20ºC. Express that temperature in the fahrenhait scale.

Answer :

we take the fahrenhait scale in question as T1 and the known Celcius scale as T2. According to equation

[5,1], we get :

+1 = �a$"a�$��a�%"a�%�a�$"a�$ + Tb1

= �$!℃"!℃��$%$℉"#$℉��%!!℃"!℃� + 32ºC

= �$!℃��%&!℉��%!!℃� + 32℉

= 36ºF+32ºF = 68ºF

2. The temperature of an iron rod with 5kg mass is going to be raised by 10 K. If the heat needed is 20

Kj, what is (a) the specific heat and (b) the heat capacity of the iron rod?

Answer :

m = 5 kg

ΔT = 10 K

Q = 20 kJ

(a) According to Equation [5-2], we get

� = ��∆a = $!,!!! �

� y���%! y� = 400 J/Kg

(b) According to Equation [5-2], we get

� = mc = �5 Kg��400 Kg. K� = 2,000 J/K

3. An object with mass of 0.5 kg at 100ºC is immersed into water with the same mass at 25ºC. If the

final temperature is 39ºC and the specific heat of the water is 4,200 J/KgºC, find the specific heat of

the object.

Answer :

mobject = mwater= 0.5 kg

ΔTobject = 100ºC-39ºC=61ºC

Cwater = 4,200 J/kgºC

ΔTwater = 39ºC-25ºC= 14ºC

The object releases heat, while the water receives heat. According to the black

Qreleased = Qreceived

(mobject)(Cobject)(ΔTobject) = (mwater)(Cwater)(ΔTwater)

Cobject = (mwater)(Cwater)(ΔTwater)

(mobject)(ΔTobject)

= (0.5 kg)(4,200 J/kgºC)(14ºC) = 963.9 J/ kgºC

(0.5 kg)(61ºC)

( NANDINI CANDRIKA )

Page 71: Soal Fisika Dan Jawaban

1) Results of measurement using water temperature is 40 ° C centigrade thermometer scale.

determine the number indicated if the water temperature was measured with a

thermometer fahrenheit scale!

Solution:

C = 40 ° C

C / 5 = (F-32) / 9

40 / 5 = (F-32) / 9 → F - 32 = 72

F = 104oF

2) If to heat 1 kg of silver which has a calorific type 230 J / kgoC. 42,000 joules of energy

needed, determine the temperature changes that occur!

Completion:

Q = m. c.∆ T

∆T = Q / (m. c) = (42,000 J) / ((1 kg) (230 J / kgoC)) = 182.6

oC

3) An object has a heat capacity of 2000 J / oC. Determine the amount of heat required to raise

the body temperature of 10 ° C!

Completion:

Given: C = 2000 J / oC

∆T = 10 ° C

Asked: Q?

Q = C. ∆T

= (2,000 J / C) (10 ° C)

= 20 000 joules = 20 kJ

4) To raise the temperature of aluminum which has a mass of 400 grams of 10 ° C to 50oC

temperature, heat is needed for 8400 J. Determine the heat of these types of aluminum!

Completion:

Given: m = 400 g = 0.4 kg

∆T = 50oC - 40

oC = 10 ° C

Q = 8400 J

Asked: c?

c = Q / (m. T)

c = (8400 J) / (0.4 kg .40o C) = 525 J / kg

oC

5) Steel bar of length 60 cm at 0oC, if the temperature is raised to 80

oC and long steel

expansion coefficient 1.1 x 10-5/oC, calculate the length of the stem after a long brick

expands and added it!

Completion:

L0 = 60 cm

t0 = 0oC

t1 = 80oC

Page 72: Soal Fisika Dan Jawaban

α = 1.1 x 10-5/oC

The length of the stem after expand are:

Lt = L0 (1 + α. ΔT)

= 60 {1 + (1.1 x 10-5

) (80-0)}

= 60 (1 + 8.8 x 10-4

)

= 60. 1,00088

Lt = 60.0528 cm

Then:

ΔL = Lt - L0

= (60.0528-60) cm

= 0.0528 cm

6) Square-shaped aluminum plate with a length of its sides is 50 cm and temperature 30oC. If

the length expansion coefficient of aluminum is 25 x 10-6

/oC, determine the broad square

aluminum if temperature is raised to 150oC!

Completion:

β0 and L0 = 50 cm

α = 25 x 10-6

/oC

A0 = β0. L0 = 2500 cm2

ΔT = 150-30 = 120oC

β = 2α = 2(25 x 10-6

/oC = 50 x 10

-6/

oC

At = Ao (1 + β. Δt)

= 2500 (1 + 50 x 10-6

/oC. 120)

= 2500 + 0.125

At = 2500, 125 cm2

7) Copper beam measuring 15 cm long, 10 cm width, and height 12 cm, has a temperature of

25oC. Copper beam is then heated to 45

oC. If the length expansion coefficient of copper 17 x

10-6

/oC, what is the volume of copper block now?

Completion:

V0 = P0. L0. T0

= 15. 10. 12 = 1800 cm3

ΔT = 45-25 = 20oC

Volume beam after heating are:

Vt = V0 (1 + 3α x Δt)

= 1800 {1 + 3 (17 x 10-6

) (20)}

= 1800 (1 + 0.00102)

= 1800 (1.00102)

Vt = 1801, 84 cm3

8) A glass volume 500 cl glass full of alcohol at a temperature of 0oC. If the glass is heated so

that the glass temperature and alcohol become 80oC, how much alcohol will spill? (Length

Page 73: Soal Fisika Dan Jawaban

expansion coefficient of glass = 0.000009 /oC and alcohol volume expansion coefficient =

0.0011 /oC)

Completion:

ΔT = 80-0 = 80oC

Glass volume: The volume of alcohol:

Vt = V0 (1 + 3α x ΔT) → Vt = V0 (1 + Υ x ΔT)

= 500 {1 + 3 (0.000009) (80)} = 500 {1 + (0.0011) (80)}

= 500 (1 + 0.000027. 80) = 500 (1 + 0.088)

= 500 (1 + 0.00218) = 500 (1.088)

= 500 (1.00218) = 544 cl

= 501.09 cl

Volume of spilled alcohol were:

ΔV = 544 - 501.09

= 42.9 cl

9) A room containing a gas volume of 6 m3 helium at a temperature of 11.75

oC. If the

temperature is raised to 80oC at constant pressure, what volume of gas now?

Completion:

Given: V0 = 6 m3

Δt = (80 - 11.75) = 68.25 oC

Vt = V0 (1 + 1 / 273 x ΔT)

Vt = 6 (1 + 1 / 273 x 68.25)

Vt = 6 (1 + 68.25 / 273)

Vt = 6 (1 + 0.25)

Vt = 6 x 1.25 = 7.5 m3

10) Heat propagate from left to right through a medium of metal rods that have a thermal

conductivity of 0.32 cal / msoC. Sectional area of the metal rod 50 cm 2, length 1 m and 40

°C temperature difference between the two ends. determine the speed of propagation of

heat in the trunk!

Completion:

k = 0.32 cal / msoC

ΔT = 40 °C

A = 50 cm2

L = 1 m

H = k. A. ΔT / L

= 0.32 cal / msoC (50 m2) ((40 C) / (1 m))

H = 0.064 cal / s

11) Find the energy emitted by a perfect black body temperature is 127oC!

Completion:

T = 127oC = 127 + 273 = 400

oK

Page 74: Soal Fisika Dan Jawaban

e = 1

σ = 5.67 x 10-8 watt/m2 (K)

4

E = e. σ. T4

= (1) (5.67 x 10-8

) (400)4

= 1452 watt/m2

12) propagate in a gas pipe with cross-sectional area 10 cm2, the gas flow from the temperature

70oC. calculate the amount of heat is transferred for 1 hour known thermal gas convection

coefficient is 0.05 cal /s m2 o

C!

Completion:

Given: A = 10 cm2 = 0.001 m

2

ΔT = 80-70 = 10 °C

Asked: Big Heat who moved?

H = h. A. Δt

= (0.05) (0.001) (10)

H = 5 x 10-4

cal/s

In 1 hour, a lot of heat that moves are:

Q = H. t

= (5 x 10-4

cal/s) (3600 s)

Q = 1.8 cal

13) Determine the amount of heat needed to heat the water as much as 2 kg of 20oC

temperature until it reaches its boiling point, which is 100oC. known to heat water type 4.2

kJkg -1

oC

-1.

Completion:

m = 2 kg

c = 4.2 kJkg-1 oC-1

ΔT = 100-20 = 80oC

Q = m. c. Δt

= (2 kg) (4.2 kJkg-1

oC

-1) (80

oC)

= 672 kJ

14) An object has a heat capacity of 2000 J oC

-1. Determine the amount of heat required to raise

temperatures of 10°C!

Completion:

Given:

C = 2000 J oC

-1

ΔT = 10 °C

Asked: Q?

Q = C. ΔT

= (2,000 J oC

-1) (10°C)

Page 75: Soal Fisika Dan Jawaban

= 20 000 joules = 20 kJ

15) Known heat melting ice 3.36 x 105Jkg

-1. Determine the amount of heat required to melt 100

grams of ice!

Completion:

Given: L = 3.36 x 105Jkg

-1

m = 0.1 kg

Q = m. L = (0.1 kg) (3.36 x 105Jkg

-1)

= 3.36 x 104 J

16) Known heat frozen water 3.36 x 105Jkg

-1. specify the amount of water which freezes when

the temperature 0oC absorbed energy of 1.68 x 10

4J of that water.

Completion: L = 3.36 x 105Jkg

-1

Q = 1.68 x 104J

M = Q / L = (1.68 x 104J) / (3.36 x 10

5Jkg

-1) = 0.05 kg / 500 grams

17) Inside there is a glass of tea as much as 60 mL of water with a temperature of 80oC. then,

into a glass of water is added 40 mL 5 oC temperature. if known heat type heat tea water

with cold water species, determine the temperature of the mixture of water!

Completion:

Given: mtea = 60 mL Ttea = 80oC

Mice = 40 mL Tice = 5oC

cteh = cair

if the final temperature of the mixture is tc, is obtained:

Q lepas (tea) = Q terima (water ice)

mteactea (Ttea - Tc) = maircair (Tc - Tair)

60 mL (80 °C - Tc) = 40 mL (Tc - 5 °C)

4800 oC - (60 Tc) = (40 Tc) - 200

oC

100 Tc = 5000 °C

Tc = 50 oC

18) The temperature scale shows the number of degrees X 70 degrees, what is the

corresponding number in degrees Celsius

Completion:

ToX = 2 / 3 (t - 10)

oC

70oX = 2 / 3 (70-10)

oC = 2 / 3 (60)

oC = 40 °C

19) What kind of heat a substance, if the mass of the object is equal to 100 grams of the

temperature will rise by 8oC when administered calories by 400 calories

Completion:

Given: Q = 400 calories

m = 100 grams

Page 76: Soal Fisika Dan Jawaban

t = 8oC

Q = m. c. Δt → c = Q / (m. Δt)

= 400 / (100 .8) = 0.5 cal / groC

20) A metal rod made of aluminum length of 2 meters at 30oC. if the length expansion

coefficient of aluminum 25 x 10-6

/ oC. what is the added length of the aluminum rod when

the temperature is raised to 50oC.

Completion:

Given: l = 2 meters

Δt = (50-30) oC = 20

oC

Δl = α. l0. Δt

Δl = (25 x 10-6

) (2) (50-30)

Δl = 10-3

m

Δl = 0.1 cm

21) Iron plate area 4 m2 at 20

oC temperature. when the temperature was raised to 100

oC, the

extent now what? expansion coefficient of long-known iron 11 x 10-6

/oC

Completion:

α = 11 x 10-6

/oC → β = 22 x 10

-6 /

oC

At = A0 (1 + α. Δt)

At = 4 {1 + 22 x 10-6

/oC (100-20)}

At = 4 (1 + 1 760 x 10-6

)

At = 4 (1 + 0.00176)

At = 4.00176 m2

22) A marble made of glass, with a long expansion coefficient of 3 x 10-6

/oC. The 2 cm diameter

marbles, at a temperature of 0oC. what is the volume of marbles that when heated to a

temperature of 100oC.

Completion:

V0 = 4 / 3 πR3 = 4 / 3 π. 1

3 = 4 / 3 π cm

2

Δt = 100oC

Υ = 3. α = 9 x 10-6

/ oC

Thus:

Vt = V0 (1 + Υ. Δt)

Vt = 4 / 3 π (1 + 9 x 10-6

. 100)

Vt = 4 / 3 π (1 + 0.0009) = 4 / 3 π + π 0.0012

Vt = 1.3345 π = 4.19 cm3

Page 77: Soal Fisika Dan Jawaban

23) Water as much as 200 grams at a temperature of 25oC is given in calories, by way of heating

at 1000 calories, if the heat kind of water 1 cal / groC. Determine the water temperature

after heating it!

Completion:

Q = m. c. Δt

1000 = 200. 1. Δt → Δt = 1000/200 = 5oC

The temperature at first t1 = 25oC

Then: Δt = t2 - t1

5 = t2 - 25 → t2 = 30oC

24) Heat capacity of a calorimeter is 100 calories. What is the change in temperature when it

absorbs heat calorimeter 840 joules.

Completion:

C = 100 calories / oC

ΔQ = 840 joules = 200 calories

remember: 1 calorie = 4.2 joules

then: C = ΔQ / Δt → Δt = ΔQ / C = 200/100 = 2oC

( RAZAN MIQDAD A. )

Page 78: Soal Fisika Dan Jawaban

KD 4.3

1. Eternity of energy [at] mixing two LOOK vitamin neglectfully kalor which [in] absorber by place of

mixing. rasher of Alumunium with mass 200 g [in] heating until its temperature reach 90 C,

later;then [is] immediately dropped into a[n containing canister 100 g irrigate [at] temperature 20 C.

Neglectfully transfer of kalor to environment [about/around] and canister absorb heat,

[count/calculate] mixture temperature. Type Kalor of alumunium 900J / singk of K, type kalor

irrigate 4200J / singk K.

ANSWER

ALUMUNIUM:

m 1= 200 g = 0,2

c1 = 200 J/kg

T1 = 90°C

AIR:

m2 = 100 g = 0,1 kg

c2 = 4200 J/kg K

T2 =20°C

Suhu akhir campuran (air dan alumunium) = x°C. x°C diantara 20°C dan 90°C

Suhu alimunium turun :

∆+1 = T1 – x

= (90 – x)°C atau (90 – x) K

Alumunium melepas kalor :

Q1 = m1c1∆+1

= (0,2 kg ) (900 J/kg K )(90 – x )

= 180 ( 90 – x ) J

Suhu air naik :

∆+2 = x – T2

= ( x – 20 )°C atau ( x -20 )

Air menerima kalor :

Page 79: Soal Fisika Dan Jawaban

Q2 = m2 c2 ∆+2

= (0,1 kg ) ( 4 200 J/kg K ( x - 20) K)

= 420 ( x - 20) J

Azaz black :

QLEPAS = QTERIMA

420 ( x – 20 ) = 180 ( 90 – x )

7 ( x – 20 ) = 3 ( 90 – x )

10 x = 270 – 3x

sX = 41

2. alloy have Mass [to] 150 gr [in] heating until 600 C, later;then into water which [is] its mass 400 temperature and gr 15 C. irrigate to stay in calorimeter which place of made him of alumunium with mass 200 gr. Balance temperature which [in] reaching [is] 28 C. Calculate type kalor of sampel alloy. ANSWER : Kalor which [in] discharging sampel = absorbed heat irrigate and calorimeter.

m 2c2 (T2-x) = m1ca ( x- T1 ) + mwcw (x-Tw)

0,1 x c2 ( 600-28) = 0 ,4 ( 4.180) ( 28- 15) + 0,2 ( 900 ) ( 28-15)

57,2 c2 = 21.736 + 2.340

57,2c2 = 24.076

C2 = 420 J/ kg°C

3. alloy have Mass [to] 150 gr [in] heating until 550 C, later;then into water which [is] its mass 400 temperature and gr 15 C. irrigate to stay in calorimeter which place of made him of alumunium with mass 200 gr. Balance temperature which [in] reaching [is] 28 C. Calculate type kalor of sampel alloy. ANSWER :

Kalor which [in] discharging sampel = absorbed heat irrigate and calorimeter

m 2c2 (T2-x) = m1ca ( x- T1 ) + mwcw (x-Tw)

0,1 x c2 ( 550-28) = 0 ,4 ( 4.180) ( 28- 15) + 0,2 ( 900 ) ( 28-15)

52,2 c2 = 21.736 + 2.340

52,2c2 = 24.076

C2 = 0,461 J/ kg°C

4. alloy have Mass [to] 150 gr [in] heating until 700 C, later;then into water which [is] its mass 400 temperature and gr 15 C. irrigate to stay in calorimeter which place of made him of alumunium with mass 200 gr. Balance temperature which [in] reaching [is] 28 C. Calculate type kalor of sampel alloy.

Page 80: Soal Fisika Dan Jawaban

ANSWER : Kalor which [in] discharging sampel = absorbed heat irrigate and calorimeter

m 2c2 (T2-x) = m1ca ( x- T1 ) + mwcw (x-Tw)

0,1 x c2 ( 700-28) = 0 ,4 ( 4.180) ( 28- 15) + 0,2 ( 900 ) ( 28-15)

67, 2c2 = 21.736 + 2.340

67,2 c2 = 24.076

C2 = 358,2 J/ kg°C

5. alloy have Mass [to] 150 gr [in] heating until 120 C, later;then into water which [is] its mass 400 temperature and gr 15 C. irrigate to stay in calorimeter which place of made him of alumunium with mass 200 gr. Balance temperature which [in] reaching [is] 28 C. Calculate type kalor of sampel alloy. ANSWER :

Kalor which [in] discharging sampel = absorbed heat irrigate and calorimeter

m 2c2 (T2-x) = m1ca ( x- T1 ) + mwcw (x-Tw)

0,1 x c2 ( 120-28) = 0 ,4 ( 4.180) ( 28- 15) + 0,2 ( 900 ) ( 28-15)

9,2c2 = 21.736 + 2.340

9,2 c2 = 24.076

C2 = 2616,9 J/ kg°C

KD 4.1

1. temperature of Tremometer mercury which not yet been calibrated to be to be plunged in ice which [is] melting ar boil, and its temperature ? can be [counted/calculated] with equation ( 6-1 ) Given Xo = 5,0 cm : X100 = 25,0 cm dan Xo = 7,8

%!! = �θ"�θ

�%!!"�%!! ; θ/100 = ), " ,!

$ ,!" ,!

%!! = 2,8/ 20 ; θ = $,& � %!! �

$! = 14 °C

2. Two A tremometer and of Bmenujukan [is] same number 100 when water nature arranging temperature irrigate moment boil. In rather warm water [of] A number menunjukan teremometer of B number menunjukan 50 . If A number menunjukan tremometer. ANSWER :

∆A : ∆B = 25 : 50

∆A : ∆B = 1 : 2

Pada skala A, QR = ∆A = 75 – 25 = 50° B

Pada skala B, QR = ∆B = ( 50 – X )° B

∆A dan ∆B 25 : 50 : 1 : 2

50 : ( 50 – X ) = 1: 2 ; 50 – X = 100 : X = -50

Page 81: Soal Fisika Dan Jawaban

3. A canister 4L, 95 [gratuity/ %] of its volume [in] alcohol content. If temperature early canister 0 and this canister [in] heating until 70 C have an effect on ( 0 ) - 1 alcohol muai koevisien 0,001 (

C°)-1.

JAWAB :

• Zat padat bejana

Vo, bejana = 4L

= 400 L

Bejana = o,oooo11 ( C° )-1

Suhu awal To = 0 °C

Akhi To = 70 °C

∆γ bejana = γ bejana Vo bejana ∆T

= 3 a bejana Vo bejana ∆T

= 3 (o, 000011 ) ( 4000 cm3) ( 70 )

= 9,24

∆γ alk = γ alk Vo alkohol ∆T

= 0,001 ( 3800 Cm3

) ( 70 ) = 266 Cm3

• Zat cair ( alkohol )

Vo alk = 95 % x 400 Cm3

γo Alk = 0,001 ( Co )-1

To = 0 °C

T = 70 ° C

Volum bejana dan alkohol suhu T = 70°C

V bejana = Vo bejana + ∆ v bejana

= 4000 + 9,24 = 4009, 24 Cm3

V alkohol = Vo alkohol + V bejana

= 4066 – 4009.24

= 56,76 Cm3

Page 82: Soal Fisika Dan Jawaban

4. Air with temperature 25 C blown to pass surface of hot plate and kalor come from bagaian in plate. Plate [is] fairish ( 40 x 40 ) temperature and cm and temperature its [is] it[him] carried through [by] [at] temperature 300 C. If coefficient transfer of convention kalor [at] mica lisp 30 W / K m2. Determining migratory kalor jumblah and [count/calculate] temperature in plate ( k = 205 g / K)

If/When made plate of alumunium thickly 1 Cm

.ANSWER :

Jumblah kalor yang berpindah :

Q = hA∆T = 30 x ( 0,4 x 0.4 ) x ( 300 – 25 ) = 1.320 joule.

Jika pelat alumunium dengan tebal 1 cm.

Q = ( k ~C) ∆T = 1320 = ( 205 )

!,%D!,!% ( T- 300 ) diperoleh T = 300,4°C

5. Water counted 2 and litre have temperature [to] 2,5 C heated in pancai till its temperature mount to become 50 C. How much/many volume irrigate after heated? ANSWER :

Vo = 2 ιdan γ = 210 x 10-6

∆T = ( 273 + 50 ) – ( 273 + 25 ) = 25 K

Jadi V = Vo ( 1= 210 x 10-6

) ( 25)

= 2,0105

( TIKA APRILIA )

Page 83: Soal Fisika Dan Jawaban

KD 4.1

1. An object has temperature of 25o C. Express the temperature of the object in the Reamur, Fahrenheit,

and Kelvin scales !

Answer

D1 : Temperature of 25o C

D2 : oF ?

oR ?

oK ?

D3 :

- In Reamur scale (oR )

C = - R → R =

- C

= - 25

o

= 20o

R

- In Fahrenheit scale (oF )

C = ` ( F – 32 ) F =

` C + 32

o

= ` 25 + 32

o

= 77 o

F

- In Kelvin scale (oK )

C = K – 273 K = C + 273

= 25 + 273

= 298 oK

2. The reading of Fahrenheit scale is equal to the Celcius scale at the temperature of

Answer

Absolute zero

3. A thermometer with oX scale has the freezing point of water at – 40

oX and boiling point of water at

160oX. if the themperature of an object is 25

oX. Determine the themperature of the object in the

Celcius, Reamur, Fahrenheit and Kelvin scales

Answer

D1 : Water freezing point -40o X

Water boiling point 160o X

Temperature of 25o

X

D2 : oF ?

oR ?

oK ?

oC ?

D3 :

- '

%!! = � "�9��

�9��"�9�m

$ %!! =

X � -!� " -!�"%D!�

25 (− 40� − 160� ) = 100 (X + 40� )

-1000X – 4000X = 100X + 4000X

-5000X = 4100X

Page 84: Soal Fisika Dan Jawaban

X = - 0,82 oC

- In Reamur scale (oR )

C = - R → R =

- C

= - . -0,82

= - 0,656 R

- In Fahrenheit scale (oF )

C = ` ( F – 32 ) F =

` C + 32

o

= ` -0,82 + 32

o

= 30,524 o

F

- In Kelvin scale (oK )

C = K – 273 K = -0,82 + 273

= 272,18 oK

4. Thermometer X which has been calibrated shows the number of -30 at water freezing point and 90 at

water boiling point. The temperature of 6o

X is equal to

Answer

D1 : Water freezing point -30

Water boiling point 90

Temperature of 6o

X

D2 : The temperature of 6o

X is equal to

D3 :

- '

%!! = � "�9��

�9��"�9�m

D�%!! =

��#! #!"`!

6� . -60 = 100 ( X + 30 )

6� . -60 = 100X + 3000

D�%!!� =

#!!!"D!

0,06X = -50

X = 80oC

KD 4.2

1. A steel wire has length of 100 cm at temperature 30oC. If the length of the steel now is 100,1

cm and = 10-5

/oC , determine the temperature of steel now !

Answer

D1 : l 100 cm = 1 m

lt 100,1 cm = 1,001 m

10-5

/oC

T1 30oC

D2 : T2 ?

Page 85: Soal Fisika Dan Jawaban

D3 :

- Lt = l0 ( 1 + . T)

1,001 = 1 ( 1 + 10-5

. T)

1,001 = 1 ( 1 + 10-5

. ∆T)

1,001 = 1 + 10-5

T

0,001= 10-5

T

T = 0,001 / 10-5

T = 100 o

C

T = T2 - T1

100 o

C= T2 – 30

T2 = 130 o

C

- Lt = l0 ( 1 + . T)

1,001 = 1 ( 1 + 10-5

. T2 - T1 )

1,001= 1 ( 1 + 10-5

. T2 – 30 )

1,001 = 1 + 10-5

. T2 – 30

1,001 – 1 = 10-5

. T2 – 30

0,001 / 10-5

= T2 – 30

100 = T2 – 30

100 + 30 = T2

130 = T2

2. A glass vessel (∝ = 9 x 10-6

/oC ) is filled with 150 cm

3 of mercury ( full ). If the coefficient of

mercury volume expansion is 1,8 x 10-4

/oC and its temperature increase 40

oC, calculate the

mercury volume which spilt out !

Answer

D1 : ∝ = 9 x 10-6

/oC

Vbejana 150 cm3

Vraksa 150 cm3

o = 1,8 x 10-4

/oC

T 0 o

C + 40 o

C = 40 o

C

D2 : Vtumpah ?

D3 :

∆Vbejana = � . V0 . T

Page 86: Soal Fisika Dan Jawaban

= 3 . 150 . 40

= 3 . 9 x 10-6

. 150 . 40

= 40500 x 10-6

= 0,0405 cm3

Vraksa = � . V0 . T

= 1,8 x 10-4

. 150 . 40

= 10800 x 10-4

= 1,08 cm3

Vbejana = V0 + Vbejana

= 150 + 0,0405

= 150, 0405 cm3

Vraksa = V0 + Vraksa

= 150 + 1,08

= 151,08 cm3

So,

Vtumpah = Vbejana - Vraksa

= 150, 0405 - 151,08

= - 1,0395 cm3

3. A gass is warmed up at constant pressure so that the temperature increase from 0 o

C to 120

oC. if the volume of gass now is 6 m

3, what is the initial volume of gass?

Answer

D1 : T1 0 o

C

T2 120 o

C

V 6 m3

D2 : V0 ?

D3 : V = V0 + T

6 = V0 + 120

V0 = 120 – 6

V0 = 114 m3

Page 87: Soal Fisika Dan Jawaban

4. A piece of metal has initial length of 10 meters at temperature 20 o

C, then is warmed up and

reaches temperature of 120 o

C. if the coefficient of length expansion is 10-4

/K, then its length

increment is

Answer

D1 : l0 10 m

T1 20 o

C + 273 =293

T2 120 o

C + 273 = 393

∝ 10-4

/K

D2 : Lt ?

D3 :

- Lt = l0 ( 1 + . T)

= 10 ( 1 + 10-4

. 100 )

= 10 ( 1 + 0,01 )

= 10 + 0,1

= 10,1 m

5. A 3 m3

gas, is increased its temperature from 12 o

C into 103 o

C at a constant pressure.

Calculate the volume of gas now !

Answer

D1 : 3 m3

T1 12 o

C

T2 103 o

C

D2 : V0 ?

D3 :

- V = V0 ( 1 + � . T )

= 3 ( 1 + %

$)# . 91 )

= 3 ( 1 + %# )

= 3 + 1

= 4 m3

KD 4.3

1. A 500 gram body absorbs heat 400 calorie so that its temperature increase 4 o

C. determine

the specific heat of that body !

Answer

D1 : Q 400 calorie

m 500 gram

T 4 o

C

D2 : c ?

Page 88: Soal Fisika Dan Jawaban

D3 :

- Q = m c T

400 = 500 . c . 4

400 = 2000 . c

c = 0,2 cal/g o

C

2. A wall has the relative constant temperature of 25 o

C in out door air temperature of 18 o

C.

calculate the dissipation heat in 3 hours because of heat convection if the wall area 15 m2 and

h = 3,5 J/s m2

K !

Answer

D1 : h = 3,5 J/s m2

K

A 15 m2

T 25

oC + 273 = 298

T 18 o

C + 273 = 291

T = 7 o

K

t 2 hours = 7200 s

D2 : Q ?

D3 :

- Q = h A T t

= 3,5 15 7 7200

= 264600 J

= 2646 x 102 J

3. A stem of 50 cm metal with section area of 20 cm2 . one of its end touches the ice

( 0 o

C ) and another end touches an object with temperature of 80 o

C. if its thermal

conductivity is 0,9 cal/s cm o

C, calculate the heat that propogate passing through the metal in

5 seconds !

Answer

D1 : d 50 cm

A 20 cm2

T 80

oC - 0

oC = 80

oC

K 0,9 cal/s cm o

C

t 5 seconds

D2 : Q ?

D3 :

- Q = H ~ ∆a

� t

= !,` $! &!

! 5

= 144 calories

Page 89: Soal Fisika Dan Jawaban

4. At temperature of 20 o

C, a certain copper with the emissivity of 0,3 area of 10 m2 � = 5,67 x

10-8

W/ m2

K-4

Answer

D1 : � = 5,67 x 10-8

W/ m2

K-4

emissivity of 0,3

T 20 o

C + 273 = 293

A 10 m2

D2 : P ?

D3 :

- P = e � A T4

= 0,3 5,67 x 10-8

10 293 293 293 293

= 125364564125.01 x 10-8

( VERONIKA PRASTIWI )

Page 90: Soal Fisika Dan Jawaban

1. At the time of ice melt thermometers x show 20 and when the boiling points number 80 . if the

temperature 40 0 C thermometer x indicates the number ...

Answer:

D1 :

Titik beku X = 200 x

Titik beku X = 800 x

Termometer celcius = 400 C

D2 :

Termometer X = … ?

D3 :

400 C = . . .

0X

# ( 40 + 20 ) = 24 + 20

= 440

x

2. Suhu nitrogen boils at 320, the Celsius scale is .... -320 ℉ = . . . ℃

`

. ( 320 – 32 )

` . -288 = 160

0

3. At what temperature centigrade thermometer and fahrenhite show the same number?

C = F

X = X

X℃ = X℉

` . X + 32 = X

32 = X . ` X

32 =( -

` ) X

32 = "- X

160 = -4 X

X = - 40 0

Page 91: Soal Fisika Dan Jawaban

4. The length of the column of mercury in the glass tube 30 mm in sushu 0℃ and 280 mm at a temperature of

100 ℃. When a long column of mercury 180 mm, then the temperature is designated. . . . . K K

h° = 30 mm

h = 280 mm

T = 100 ℃

T° = 0℃

180 – 30 = 280 – 30 = Panjang

X – 0 100 – 0 Suhu

= 150 = 250

X 100

X = 15000 = 15 . 4 = 60 ℃

250

• 60 + 273 = 333 ° K

5. The temperature in an air-conditioned room is 68. In Celsius temperature scale is shown. . . . .

` ( 68 - 32 ) =

` x 36 = 20

6. A new type of liquid with a mass of 500 grams and the temperature 10 ℃. The substance is heated

to 50 ℃. Turned out to require 2800 J of heat. Heat type material was ...

M = 500 gr = 0,5 kg

T1 = 10 ℃

T2 = 50 ℃

Q = 2800 J

C … ?

Q = M . C . ∆t

2800 = 0,5 . c . 40

C = 2800

0,5 . 40

C = 2800

20

C = 140 J/kg℃

Page 92: Soal Fisika Dan Jawaban

7. Steel and brass rod cross-sectional area and panjangna same, one end is connected. suhu ujung

batang bajanya yang bebab 250 . sedangkan suhu ujung batang kuningan yang bebab 100 . jika

koefisien masing – masing 0,12 kal/s dan 0,24 kal/s maka suhu pada titik hubung kedua batang

tersebut adalah … temperature of the steel rod tip which bebab 250. while the brass rod tip

temperature bebab 100. if the coefficient of each - each 0.12 cal / s and 0.24 cal / s, the

temperature at the point connecting the two rods are ...

K1 . A1 . ∆£ = K2 . A2 . ∆£

L L

0,12 ( 250 – tc ) = 0,24 ( tc – 24 )

30 – 0,12tc = 0,24 tc – 24

30 + 24 = 0,24 tc + 0,12 tc

54 = 0,36 tc

-

!,#D = 150 ℃

8. Skin temperature of a person think - about 32 ℃ . If people think outside the body surface - about 1.6 m 2

are in the room temperature is 22 ℃ , then the person's body heat is released through convection for 5

minutes is. . . . .

¤¥ = H . A . ∆£

T1 = 32 ℃ = 32 + 273 = 305 °K

A = 1,6 m2

T2 = 22 ℃ = 22 = 273 °K

t = 5 menit = ( 5 x 60 ) 300 detik

H = 77 W/mK

Q ..?

B

#!! = 77 . 1,6 . -10

B

#!! = -1232

Q = -1232 . 300 = - 369600

• Min menyatakan menyerap panas .

9. A thermometer with a scale-free x has a freezing point of water at - 40 x and 160 x water point. At the time

of the thermometer read 15 x then the Celsius scale thermometer reads ...

Xmin = -40 °x

Xmax = 160 °�

X = 15 °x

Page 93: Soal Fisika Dan Jawaban

℃ = . . . ?

Zdee =

­"­S®Q­SP¯"­S®Q

'%!! =

% "�"-!�%D!"�"-!�

'%!! =

$!!

C = $ = 27,5 ℃

10. Thermometer reamur and Fahrenheit thermometer will show the numbers in the measurement of

temperature. . . . .

T°\ = T°�

R = `

- R + 32

-

- R -

`

- R = 32

"

- R = 32

R = 32 X ("-

)

R = -25,6°\

11. If 75 grams of water where the temperature 0℃ is mixed with 50 g of water temperature is 100 ℃ then it

is a mixture of water temperature. . . . .M1 = 75 gr

T1 = 0℃

M2 = 50 gr

T2 = 100℃

Tc … ?

QLEPAS = QTERIMA

M2 . C2 . ∆=2 = M1 . C1 . ∆=1

50 ( 100 – tc ) = 75 ( tc – 0 )

5000 – 50 tc = 75 tc

5000 = 125 tc

!!!

%$ = 40℃ = tc

12. When the heat kind of ice = 0.5 cal / g is to raise the temperature of 800 grams of ice from - 12 to 60℃ it

takes the heat as much ... (if in the heat of ice 80 cal / g and the heat kind of water 1 cal / g)M = 800 gr

Ces = 0,5 kal/gr℃

L = 80 kal/gr

Cair = 1 kal/gr℃

Page 94: Soal Fisika Dan Jawaban

T1 = -12 ℃

T2 = 60℃

Q1 = 800 . 0,5 ( 0 –(-12))

= 400 . 12

= 4800 kal

Q2 = 800 . 80

= 64000 kalori

Q3 = 800 . 1 ( 60-0 )

= 800 . 60

= 48000 kal

Qtotal = Q1 + Q2 + Q3

= 4800 + 64000 + 4800

= 116.800 kalorii

13. A bottle made of glass volume 1000 cm 3. alcohol bottles filled to the brim, and then heated dare

30-90 � hglass = 9 x 10 -6 ℃ oalcohol = 10

-3 / ℃�

Tentukan volume alkohol yang tumpah . Determine the volume of spilled alcohol.

Vbotol = 1000 cm3

Valkohol = 1000 cm3

∆= = 60℃

hkaca = 9 x 10-6

/℃

oalkohol = 10-3

/℃

Vtumpah . . . ?

VT alkohol = V0 ( 1 + o .∆= )

= 1000 ( 1 + 10-2

. 6 )

= 1000 + 60 = 1060 cm3

VT botol = V0 ( 1 + o .∆= )

= 1000 ( 1 + 3 . 9 . 10-6

. 6 . 10-1

)

= 1000 + 162 . 10-2

= 1000 + 1,62 = 1001 , 62 cm3

∆± tumpah = VT botol – VT alkohol

= 1001,62 – 1060

= - 58 , 38 cm3

( YULI LESTARI )

Page 95: Soal Fisika Dan Jawaban

1. Determine calor required just for changes 1 kg ice gets temperature -20oC . as

waters the lot on pressure 1 atm gets temperature 50oC , if calor ice type 0,5

kkal / kgoc , calor melts ice 80 kkal / kgoc . And calor water type 15 = 1 kkal /

kgoc.

Answer :

� D1 = mice = 1 kg

= Tice = -20 oC

= Twater = 50oC

=Cice =0,5 kkal / kgoc

=Cwater = 1 kkal / kgoc

= Lice = 80 kkal / kgoc

� D2 = Q ...... ?

� D3 = Q1 = mice . Cice .∆+ = 1 . 0,5 . 20

= 10 kkal

= Q2 = mice . Lice

=1 . 80

= 80 kkal

= Q3 = mwater . Cice . ∆+

= 1 . 1. 50

= 50 kkal

= .∆> = Q1 . Q2 . Q3

= 10 + 80 + 50

= 140 kkal

2. If 400 g ice get temperatures -10oC be interested into 300 g water . Get

temperature 20oC , Meanwhile calor’s interchange just among ice and even water

, calors melts ice 80 kkal / g . Calor ice type 15 0,5 kal / g . Calor is water type 1

kal / goC . How much temperature in a state that mixture final ?

Answer :

� D1 = mice = 400 g

= mwater = 300 g

= Tice = -10 oC

= Twater = 20oC

=Cice =0,5 kkal / kgoc

Page 96: Soal Fisika Dan Jawaban

=Cwater = 1 kkal / kgoc

= Lice = 80 kkal / kgoc

� D2 = t ....... ?

� D3 = Q1 = mice . Cice .∆+ = 400 . 0,5 . 10

= 200 kkal

= Q2 = mice . Lice

=400 . 80

= 3200 kkal

= Q3 = mwater . Cice . ∆+

= 300 . 1. 20

= 6000 kkal

= m = B#"B%

B$ . S²@³

= D!!!"$!!!

#$!! .400 = 50 A

3. An object’s temperature changes from 70oF into 98oF . State those

temperature in the celcius , reamure , and Kelvin scale .

Answer :

� -28oF = oC

=` − 28 = "$ $�#$

= "$$! = −44

� -28oF = oR

=`- − 28 = −63

� -28oF = oK

= - 28 + 273 = 245

4. The temperature of an iron rod with 5 kg mass is going to be raised by 10 K . If

the heat needed is 20 kj , how much is the specific heat and heat capacity of

the iron rod ?

Answer :

� m = 5 kg

T = 10 K

Page 97: Soal Fisika Dan Jawaban

Q = 20 kj

� (a) =C = B

9 ∆Y = $!,!!! G� HI ��%! ( � = 400 J/kg.K

� (b) = C = (5 kg )(400 J/kg.K)

= 2,000 J/K

5. How much out grow if calor that at needs to thaw ice as much 100 g , on

temperature 0oC as water the lot that 0oC ? if at knows calor laten ice fusion

becomes water as big as 80 kal / gr ?

Answer :

� D1 = m = 100 gr.

L = 80 kal/gr

� D2 = Q……..?

� D3 = Q = m . L

= 100 .80

=8000 kal

6. One object with calor’s capacity 100 Joule / oC . How much calor one at need to

raise that object temperature as big as 60oC ?

Answer :

� D1 = C = 100 Joule / oC

=∆+ = 60oC

� D2 = Q ……?

� D3 = Q = c . ∆+

= 100 . 60

=6000 Joule

( ZUMROTUL ULYA )

Page 98: Soal Fisika Dan Jawaban

1. The temperature of an object expressed in the Celsius scale is 30oC. State theobject temperature in Kelvin scale!

SOLUTION

We have celcius scale is 300C

• ay – $)# #)#"$)# =

#!"! %!!"!

• ay – $)#

%!! = #! %!!

Tk = �30 + 273�

= 303 oC

( Muhammad fatkhul arifian )


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