HEAT EXCHANGE

The exchange of thermal energy is simply referred to as heat.

If an object rises/falls in temperature it has gained/lost thermal energy.Heat from here on in refers to the exchange of thermal energy

therefore, heat refers to either “warming up” or “cooling down”

How much heat energy an object can gain or lose depends upon 3 factors:

1) Mass

2 kg of iron

1 kg of iron

2) Temperature Change

3) Type of Substance1 kg of Al1 kg of Fe

Heat Exchange

Different objects will gain/lose thermal energy at different rates:

• 1 kg of aluminum will melt more ice than 1 kg of iron

• 1 kg of iron will melt more ice than 1 kg of lead

The amount of heat that an object will absorb in order to raise its temperature 1˚C is called the heat capacity of the object.

Heat Capacity refers to how much heat an object can gain/lose.

• 1 kg of Al will have a higher heat capacity than .5 kg of Al because of the larger mass even though they gain and lose heat at the same rate.

Specific Heat (c) is the amount of heat needed to raise 1 g of a substance 1˚C:

c = Q

m∆T

units:

J

g˚C

Specific heats at 1 atm :

Substance c [J/(g˚C)]

Aluminum .900

Copper .390

Iron (steel) .450

Lead .130

Alcohol 2.40

Water 4.19

Ice 2.10

Using this definition, the amount of heat needed to warm/cool a substance can be calculated:

Q = mc∆T

units: J

The Law of Heat Exchange:

When 2 or more objects of different temperatures are mixed together, they will exchange heat until reaching a final, equilibrium temperature.Q lost = Q

gained

A copper calorimeter with a mass of 150.0 g contains 350.0 g of water. The temperature of both the calorimeter and the water is 20.0˚C. A metal cylinder, with a mass of 200.0 g and a temperature of 99.5˚C, is placed into the calorimeter. The final temperature of the calorimeter, water, and cylinder is 26.7˚C. Find the specific heat of the metal cylinder.

mc = 150.0 gmw = 350.0 gTc/w = 20.0˚C

mm =200.0 g

Tm = 99.5˚C

Tc/w/m = 26.7˚C

cm = ?

cc = .387 J/g˚Ccw= 4.19 J/g˚C

Qlost = Qgained

(mc∆T)m =

(mc∆T)c + (mc∆T)w

(200.0g)(cm)(72.8˚C)

= (150.0g)(.387 J/g˚C)(6.7˚C)+ (350.0 g)(4.19 J/g˚C)(6.7 ˚C)

14,560(cm) = 10,214

cm = .702 J/g˚C

A 115 g piece of copper is heated to 99.8ºC and then placed in a 65.0 g Aluminum calorimeter that contains 200.0 g of water at 13.5ºC. What will the equilibrium temperature be?mc= 115gcc = .387 J/gºCTc = 99.8ºC

ma = 65.0 g

ca = .909 J/gºC

mw = 200.0 g

cw = 4.19 J/gºC

Ta/w = 13.5 ºC

T = ?

NOTE: Temperature changes in heat exchange problems are always Higher minus Lower!

Qlost = Qgained

(mcT)c= (mcT)w + (mcT)a

(115 g)(.387 J/g˚C)(99.8 - T) =(200.0 g)(4.19 J/g˚C)(T - 13.5) +

(65.0 g)(.909 J/g˚C)(T - 13.5)

4440 - 44.5(T)=

838(T) - 11,300 +

59.1(T) - 798

16,538 = 942(T)

T = 17.6 ˚C

1) A 75.0 g aluminum calorimeter contains 100.0 g of water at 20.0 ˚C. A student wishes to heat the water to 100.0 ˚C by immersing a piece of iron heated in a furnace to 350.0 ˚C. How much iron will he need to use? 2) What will be the equilibrium temperature when a 125 g copper block at 99.0 ˚C is dropped into a 50.0 g aluminum calorimeter containing 145 g of water at 15.0 ˚C?3) A 170.0 g copper calorimeter containing water at 19.5 ˚C is raised to 42.1 ˚C when 75.0 g of water at 90.5 ˚C is poured in. How much cool water was originally in the cup?

Heat of Fusion/Vaporization

When an object gains/loses heat, one of two things must happen:

1) It changes temperature.

OR2) It changes phase.

It cannot do both- it is one or the other!When a substance gain/loses

heat without changing temperature, it must be

changing potential energy and therefore changing phase!

When a substance changes phase, it must gain or lose heat to do so it stores or gives up stored energy: The amount of heat needed to melt/freeze one gram of a substance is called the Heat of Fusion (Latent Heat-- Lf):Lf

=

Q

mQ = mLf

It requires 334 J to turn one gram of ice at 0˚C into water at 0˚C. For water: Lf = 334

J/g

The amount needed to boil (vaporize) one gram of a substance is called the Heat of Vaporization (Lv):

Lv =Q

m Q = mLv

How much heat would be needed to raise 10.0 g of ice at -20.0˚C and turn it into steam?

For water: Lv = 2260 J/g

1) “warm” the ice:

Q=mc∆T

=(10.0g)(2.22 J/g˚C)(20.0˚C)= 444 J

2) melt the ice:Q = mLf

= (10.0 g)(334 J/g) = 3340 J

3) warm the water:Q = mc∆T

= (10.0 g)(4.19 J/g˚C)(100.0˚C) = 4190 J4) boil the water: Q = mLv

= (10.0 g)(2260 J/g) = 22,600 J

The total amount of heat energy needed:444 J + 3340 J + 4190 J + 22,600 J = 30,

574 JApproximately 74% of the thermal energy gained is acquired while changing phase from liquid to gas. This is why steam burns so much worse than boiling water even though they are both examples of H2O at 100˚C!

An aluminum calorimeter cup of 50.0 g is filled with 120.0 g of water at 5.0 ˚C when 11.7 g of steam is pumped into the calorimeter. The final temperature of the calorimeter, water and condensed steam is measured to be 55.0 ˚C. What is the experimental heat of vaporization of water for this trial?

mc = 50.0 gcc = .909 J/(g˚C)mw = 120.0 g

cw = 4.19 J/(g˚C)Tw+c = 5.0˚ C

ms = 11.7 gT = 55.0 ˚ C

Lv = ?

An aluminum calorimeter, mass of 150.0 g, contains 420.0 g of water at 35.0˚C. An 83.5 g amount of ice at 0˚C is placed into the calorimeter and it melts completely. The final temperature is 17.0 ˚C. Calculate the heat of fusion for water.

ca = .909 J/g˚Cma = 150.0 g

mw = 420.0 g

cw = 4.19 J/g˚C

mi = 83.5 gTf = 17.0˚C

Lf = ?

Ta+w = 35.0˚C

Ti = 0

Qlost = Qgained

(mc∆T)w + (mc∆T)a= miLf + micw∆T

(420.0 g)(4.19 J/g˚C)(18.0˚C)+ (150.0 g)(.909 J/g˚C)

(18.0˚C) =(83.5 g)(Lf) + (83.5 g)(4.19 J/g˚C)(17.0˚C)

31,700 + 2450 = 83.5Lf + 5950

Lf = 28,200

83.5

= 338 J/g

What is the final temperature when 25.0 g of steam is pumped into a 150.0 g aluminum calorimeter that contains 200.0 g of water at 35.0˚C?Tf = ?

ms = 25.0 g

Lv = 2260 J/gma = 150.0 gca = .909

J/g˚Cmw = 200.0 g

cw = 4.19 J/g˚CTw+a = 35.0˚C

Remember: When steam (or ice) is involved you must account for heat gained/lost while changing phase!

Q lost = Q gained

msLv + mscw∆T =maca∆T + mwcw∆T

(25.0 g)(2260 J/g) +

(25.0 g)(4.19 J/g˚C)(100˚C - Tf)= (150.0 g)(.909 J/g˚C)(Tf

- 35.0˚C)+ (200.0 g)(4.19 J/g˚C)(Tf - 35.0˚C)

56,500 + 10,500 - 105 Tf

= 136 Tf - 4770 + 838 Tf - 29,300

101,000 = 1080 Tf

Tf = 93.5 ˚C