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:
2 kg of iron
1) Mass
1 kg of iron
2) Temperature Change
1 kg of Al
3) Type of Substance
1 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 g
mw = 350.0 g
Tc/w = 20.0˚C
mm =200.0 g
cm = ?
Tm = 99.5˚C
cc = .387 J/g˚C
Tc/w/m = 26.7˚C
cw= 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= 115g
mw = 200.0 g
cc = .387 J/gºC
cw = 4.19 J/gºC
Tc = 99.8ºC
Ta/w = 13.5 ºC
ma = 65.0 g
T=?
ca = .909 J/gºC
NOTE: Temperature changes in heat exchange
problems are always Higher minus Lower!
Qlost = Qgained
(mcT)c= (mcT)w + (mcT)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.
OR
2) 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):
Q
Lf =
Q = mLf
m
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
For water: Lv = 2260 J/g
How much heat would be needed to
raise 10.0 g of ice at -20.0˚C and turn
it into steam?
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 J
4) 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 J
Approximately 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?
cw = 4.19 J/(g˚C)
mc = 50.0 g
cc = .909 J/(g˚C)
mw = 120.0 g
Tw+c = 5.0˚ C
ms = 11.7 g
T = 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˚C
Ta+w = 35.0˚C
ma = 150.0 g
Ti = 0
mi = 83.5 g
mw = 420.0 g
cw = 4.19 J/g˚C
Tf = 17.0˚C
Lf = ?
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/g
ma = 150.0 g
ca = .909 J/g˚C
mw = 200.0 g
cw = 4.19 J/g˚C
Tw+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
A copper ball of mass 4.54 kg is removed from a
furnace and dropped into a bucket with 1.36 kg of
water at 22.0˚C. After the water stops boiling, the
combined mass of the water and the copper ball is
5.45 kg. What was the temperature of the furnace?
(No heat was absorbed by the bucket.)
A mixture of ice and water, mass 200.0 g, is in a
100.0 g calorimeter with a specific heat of .800
J/(g˚C). When 40.0 g of steam is pumped into that
mixture, the final temperature is 60.0˚C. How
many grams of ice were in the calorimeter
originally?