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Calorimetry is that branch of heat which deals withits measurement. the useual units of heat are calorieor kilocalorie

1. SPECIFIC HEAT

This is also called ' Heat Inertia' of a substance

1.1 Specific Heat of solid & Liquid (s) : If a solid or aliquid is heated till the change of the state, there isno change in their volume hence the work done iszero. Hence there is only one specific heat of solidsand liquids which is constant.

1. Definition : The amount of heat needed for anunit increase in the temperature of unit mass ofa solid or liquid is called it's specific heat

Unit : kilocalorie / kg-ºC or calorie/g-ºC

1 Kcal / Kg-ºC = 1 Cal/g-ºC

2. If mass of the body is 'm' and specific heat is 's'then amount of heat needed to increases it'stemperature by dT is given by Q = msdT

3. Specific heat of water = 1 kcal / kg-ºC = 1Cal/ g-ºC = 4.18 × 103 J/kg-ºC

4. Kelvin can also be used instead of ºC is size ofboth uints is same.

1.2 Specific heat of gas (C)

(1) There are many processes possible to give heatto a gas. A specific heat can be associated toeach such process which depends on the natureof process.

(2) The number of possible specific heats for a gasis infinite and the value of specific heats can

very from zero (0) to infinity ().

(3) Generally two types of specific heat arementioned for a gas -

(a) specific heat at constant volume (Cv)

(b) specific heat at constant pressure (Cp)

(4) These specific heats can be molar or gram,depending on the amount of gas considered todefine it

(5) The molar heat capacities of a gas are defined asthe heat given per mole of the gas per unit risein the temperature

(6) The molar heat capacity at constant volume,

denoted by Cv is Cv =

Tn

Q

(7) If, at absolute temperature T , total energy of agas E , degree of freedom of gas = f

E =2

fRTand Cv =

T

E

CV =2

Rf

(8) Amount of heat needed to increase thetemperature of n moles of gas by dT at constantvolume is

d = ncv dT.

(9) Amount of heat needed to increase thetemperature of 1gm of gas by 10C at constantvolume is called gram specific heat at constantvolume.

(10) (Cv)gram =M

)C( molarv

or (Cv)molar = M(Cv)gram

where M = molecules wt . of gas

(11) Amount of heat needed to increases temperatureof 'm' gm of gas by dT at constant volume is

dQ = m (Cv)gram dT =M

m(Cv)molar dT

(12) Amount of heat needed to increase temperatureof 1mole of gas by 10C at constant pressure iscalled molar specific heat at constant pressure.

(13) Cp = Cv + R, for one mole.

where R = universal gas constant

(14) Amount of heat needed to increase temperatureof n moles of gas by dT is at constant pressure

dQ = nCp dT

Note : If gas is heated at constant pressure, then Cv canbe replaced by Cp in above discussion. Hence

(a) (Cp)gram =M

CP

or Cp = M (Cp)gram

(15) Specific heat for other processes

Tm

QS

(a) adiabatic s = 0, as Q = 0 but T has somevalue

(b) isothermal s = , as T = 0 but Q has somevalue

(16) Specific heat for any process is given by

C = Cv +ndT

pdV= Cp – R +

ndT

pdV

(for the n mole of gas)

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Example :

(1) If process is isochoric , then dV = 0

C = Cv

(2) If process is isobaric , then pV = nRT

dT

pdV= nR

C = Cv + R = Cp (for one mole)

2 THERMAL CAPACITY

(1) Amount of heat needed to increase thetemperature of a substance (any amount) by 1ºCis called thermal capacity of that substance.

(2) Thermal capacity = ( mass of body ) x (specific

heat) Hc = ms

(3) Unit = calorie /ºCor Kcal/ºC

Heat capacity at point 'p' =tan

1= cot

(4) Thermal capacity is given by reciprocal of slopeof heat temperature curve.as

Hc = mS =dT

Q

(5) Heat capacity in an isothermal process is infinite

(). e.g. process of melting and vaporisation

(6) If heat capacity of a body is Hc, then heat needed

to rise it's temperature by d is, dQ = Hc d

3. WATER EQUIVALENT OF A BODY

(1) If m gram of a substance is given Q amount ofheat which rises its temperature by T . Now ifon giving same amount of heat temperature of wgram of water is also increased by T then w iscalled water equivalent of body of mass m.

(2) The value of water equivalent of a body is sameas it's heat capacity. The difference is only inunits. e.g If heat capacity of a body is m calorie/0C then it's water equivalent will be m gram.

(3) Physical meaning : The same amount of heathas to be given to a body for increasingit'stemperature by dT as needed for quantity ofwater equal to it's water equivalent by sametemperature range.

4. LATENT HEAT

Latent heat of fusion of a substance is the quantityof heat (in kilocalories) required to change its 1 kgmass from solid to liquid state at its melting point(For ice latent heat of fusion = 80 kilocal/kg).

Latent heat of vaporization of a substnce is thequantity of heat required to change its 1 kg massfrom liquid to vapour state at its boiling point.

For water latent heat of vaporisation = 536 kilocal/kg

Specific heat, thermal capacity &latent heat

Ex.1 The amount of heat necessary to raise thetemperature of 0.2 mol of N2 at constant pressurefrom 37º to 337º C will be -

(A) 1764 Joule (B) 764 Joule

(C) 1764 Calorie (D) 1764 erg

Sol.[A] dQ = nCPdT

dQ = 0.2 × 7 × 300

dQ = 420 Calorie

dW = 420 × 4.2

dW = 1764 Joule

Ex.2 (Cp – Cv) will be -

(A) Maximum for oxygen

(B) Maximum for nitrogen

(C) Maximum for carbon di oxide

(D) Same for all gases

Sol.[D] (Cp – Cv) = R, where R is gas constant which issame for all gases. So (Cp – Cv) is same for allgases.

Ex.3 The ratio of radii of two copper spheres is2 : 1 and they are kept at same temperature. Theratio of their heat capacities will be -

(A) 2 : 1 (B) 1 : 1

(C) 8 : 1 (D) 4 : 1

Sol.[C]dQ

dQ1

2=

4343

13

23

r S

r S=

r

r1

2

3FHGIKJ =

2

1

3FHGIKJ =

8

1

Ex.4 5 g of ice at 0°C is dropped ina beaker containing20 g of water at 40°C, then

(A) All the ice will not melt into water

(B) All the ice will melt and the resultingtemperature of water will be 0°C

(C) All the ice will melt and the resultingtemperature of water will be 25°C

(D) All the ice will melt and the resultingtemperature of water will be 16°C

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Sol.[D] Heat required to melt 5 g ice = 5 × 80 = 400 cal. Heatavailable with 20 g water at 40°C = 20 × 1× 40= 800 cal out of this, 400 cal heat will be used tomelt the ice and remaining heat will raise thetemperature. Thus Heat given by water = Heattaken by ice

20 × 1 × (40 – ) = 5 × 80 + 5 × 1 × ( – 0)800 – 20 = 400 + 5

25 = 400

or =25

400=16°C

5. GASEOUS MIXTURE

(i) Suppose two non - reactive gases are enclosedin a vessel of volume V. In the mixture n1 molesof one gas are mixed with n2 moles of anothergas. If NA is Avogadro's number then number ofmolecules of first gas will be N1 = n1 NA andnumber of molecules of second gas will beN2 = n2A2.

(ii) Total mole fraction n = (n1 + n2).

(iii) If M1 is molecular weight of first gas and M2that of second gas the molecular weight of themixture will be

M =n M n M

n n1 1 2 2

1 2

(iv) Specific heat of the mixture at constant volumewill be

Cv =n C n c

n n

v v1 2

1 2

1 2

(v) Specific heat of the mixture at constant pressurewill be

Cp =n C n c

n n

p p1 2

1 2

1 2

(vi) Ratio of specific heats of the mixture will be

=C

Cp

v =

n C n C

n C n Cp p

v v

1 2

1 2

1 2

1 2

(vii) Pressure exerted by the mixture

P = (n1 + n2)RT

V

=m

M

m

M1

1

2

2

FHG

IKJ

RT

V(where m1 and m2 are the masses of the twogases respectively)

(viii) If n1 moles of first gas at a temperarure T1 aremixed with n2 moles of the other gas at atemperature T2, the temperature of the mixturewill be

T =n T n T

n n1 1 2 2

1 2

Gaseous mixture

Ex.5 n1and n2 moles of two ideal gases of thethermodynamics constants 1 and 2 respectively

are mixedC

CP

v for the mixture is -

(A) 1 2

2

(B)n n

n n1 1 2 2

1 2

(C)n n

n n1 2 2 1

1 2

(D)n n

n n1 1 2 2 2 1

1 2 2 1

1 1

1 1

( ) ( )

( ) ( )

Sol.[D] At constant temperature U = U1 + U2

nCv = n1 Cv1 + n2 Cv2

Total number of mols. in the mixture

n = n1 + n2

nCv = n1R

1 1 + n2

R

2 1 CP = Cv

nCP = n1 1R

1 1 + n2 2

R

2 1

=C

CP

v =

nR

nR

nR

nR

1 11

2 22

11

12

1 1

1 1

=n n

n n1 1 2 2 2 1

1 2 2 1

1 1

1 1

( ) ( )

( ) ( )

Ex.6 One mole of monatomic gas ( = 5/3) is mixedwith one mole of a diatomic gas ( = 7/3), thevalue of for the mixture is

(A) 1.40 (B) 1.50

(C) 1.53 (D) 3.07

Sol.[C] =21

21

V2V1

P2P1

CnCn

CnCn

=

23

x125

x1

25

x127

x1

= 1.5

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SOLVED EXAMPLEEx.1 1 g of steam at 100°C can melt how much ice at

0°C? Latent heat of ice = 80 cal/g and latent heatof steam = 540 cal/g.

Sol. Heat required by ice for melting of m g of ice =mL = m × 80 cal

Heat available with steam for being condensedand then brought to 0°C

= 1 × 540 × 100

= 640 cal

m × 80 = 640

or m = grams880

640

Ex. 2 A tap supplies water at 10°C and another tap at100°C. How much hot water must be taken sothat we get 20 kg of water at 35°C ?

Sol. Let mass of hot water = m kg

mass of cold water

= (20 – m) kgHeat taken by cold water

= (20 – m) × 1 × (35 – 10)Heat given by hot water

= m × 1 × (100 –35)Law of mixture gives

Heat given by hot water

= Heat taken by cold water

m × 1 × (100 – 35) = (20 – m) × (35 – 10)65 m = (20 – m) × 2565 m = 500 – 25 m

or 90 m = 500

m =90

500= 5.56 kg

Ex. 3 5 g of ice at 0°C is dropped in a beaker containing20 g of water at 40°C. What will be the finaltemperature?

Sol. Let final temperatue be =

Heat taken by ice = m1L + m1c1 1

= 5 × 80 + 5 × 1 ( – 0)

= 400 + 5 Heat given by water at 40°C

= m2c2 2 = 20 × 1 × (40 – )

= 800 – 20 As Heat given = Heat taken

800 – 20 = 400 + 5

20 = 400

=25

400 = 16° C

Ex. 4 5 g ice of 0°C is mixed with 5 g of steam at 100°C.What is the final temperature?

Sol. Heat required by ice to raise its temperature to100°C,

Q1 = m1L1 + m1c1 1

= 5 × 80 + 5 × 1 × 100

= 400 + 500 = 900 cal

Heat given by steam when condensed,

Q2 = m2L2

= 5 × 536 = 2680 cal

As Q2 > Q1. This means that whole steam is noteven condensed.

Hense temperature of mixture will remain at 100°C