THE QUANTITY OF HEAT
The thermal energy lost or gained by
objects is called heat.
One calorie (cal) is the quantity of heat
required to change the temperature of one
gram of water through one Celsius degree.
The mechanical equivalent of heat is given
by: 1 cal = 4.186 J
SPECIFIC HEAT CAPACITY
The amount of thermal energy required to raise the
temperature of a substance varies for different materials.
The specific heat capacity of a material is the quantity of
heat required to raise the temperature of a unit mass
through one degree.
Q = m c T
Units:
Q in J (cal)
m in kg (g),
T in K (ºC),
c in J/kg.K (cal/g ºC)
Specific Heat Capacity
Same heat is absorbed.
Iron's ability to store heat is less than water's.
Iron's temperature rises more than does the water's.
14.1 How much heat is required to raise the temperature of
200 g of Hg from 20C to 100C?
m = 200 g
to = 20C, tf = 100C
c = 0.033 cal/gC
Q = m c T
Q = 200 (0.033) (100 - 20) = 528 cal
MEASUREMENT OF HEAT
The principle of thermal equilibrium is the result of a transfer
of thermal energy from the warmer bodies to the cooler bodies.
If energy is to be conserved we say that the heat lost by the
warm bodies must equal the heat gained by the cool bodies.
14. 2 A handful of copper shot is heated to 90C and then
dropped into 80 g of water at 10C. The final temperature of
the mixture is 18C. What was the mass of the shot?
mw = 80 g, cw = 1 cal/gC, tw= 10C
teq = 18C
Q LOST(Cu) = Q GAINED (water)
tCu = 90C, cCu = 0.093 cal/gC
mCu cCu (tCu – teq ) = mw cw (teq – tw )
mCu 
m w c w ( t eq  t w )
c C u ( t C u  t eq )
mCu 
( 80 )(1)(18  10 )
( 0.093)( 90  18 )
= 95.5 g
In this example we have neglected two important facts:
o the water must have a container which will absorb heat
from the shot
o the entire system must be insulated from external
temperatures otherwise, the equilibrium temperature
will always be room temperature
A laboratory device called a calorimeter is used to control these
difficulties.
14.3 80 g of dry iron shot is placed in a cup and heated to a
temperature of 95C. The mass of the inner aluminum cup and
of the aluminum stirrer is 60 g. The calorimeter is partially
filled with 150 g of water at 18C. The final temperature of the
system is 22C. Find the specific heat capacity of iron.
mFe = 80 g, tFe = 95C
mAl = 60 g, cAl = 0.22 cal/gC
mw = 150 g, cw = 1 cal/gC, tw= 18C
teq = 22C
Q LOST(Fe) = Q GAINED (water) + Q GAINED (Al)
QGAINED:
Qw = (150)(1)(22-18) = 600 cal
QAl = (60)(0.22)(22-18) = 52.8 cal
Q G(w+Al) = 652.8 cal = Q L(Fe) = mFe cFe T
c Fe 
c Fe 
Q G A IN E D
m Fe  t
652 .8
( 80 )( 95  22 )
= 0.11 cal/gC
CHANGE OF PHASE
The change of phase from a solid to a liquid is called fusion,
and the temperature at which this change occurs is called the
melting point of the substance.
The latent heat of fusion Lf of a substance is the heat per unit
mass required to change the substance from the solid to the
liquid phase at its melting temperature.
Q = m Lf
The change of phase from a liquid to a vapor is called
vaporization, and the temperature at which this change occurs
is called the boiling point of the substance.
The latent heat of vaporization Lv of a substance is the heat per
unit mass required to change the substance from a liquid to a
vapor phase at its boiling temperature.
Q = m Lv
The change of phase from a vapor to a liquid is called
condensation.
heat of condensation = heat of vaporization
The change of phase from liquid to solid is called freezing or
solidification.
heat of solidification = heat of fusion
The change of phase from solid to gas without passing
through the liquid state is called sublimation.
CHANGES OF PHASE
TEMPERATURE vs. QUANTITY OF HEAT
14.4 What quantity of heat is required to change 20 g of ice at
-12C to steam at 100C?
mice = 20 g, cice = 0.5 cal/gC, tice = - 12C
cw = 1 cal/gC
tsteam = 100C
Lf = 80 cal/g, Lv = 540 cal/g
Q1: heat needed to raise the tice to its melting point
Q1 = m c T =20(0.5)(0-(-12)) = 120 cal
Q2: heat needed to completely melt the ice
Q2 = m Lf = 20(80) = 1600 cal
mice = 20 g, cice = 0.5 cal/gC, tice = - 12C
cw = 1 cal/gC
tsteam = 100C
Lf = 80 cal/g, Lv = 540 cal/g
Q1: heat needed to raise the tice to its melting point
Q1 = m c T =20(0.5)(0-(-12)) = 120 cal
Q2: heat needed to completely melt the ice
Q2 = m Lf = 20(80) = 1600 cal
Q3: heat needed to raise the tw to its boiling point
Q3 = m c T =20(1)(100-0) = 2000 cal
Q4: heat needed to completely vaporize the water
Q4 = m Lv = 20(540) = 10,800 cal
QTOTAL = 120 + 1600 + 2000 + 10800
= 14,520 cal
14.5 After 12 g of crushed ice at -10C is dropped into a 50 g
aluminum calorimeter cup containing 100 g of water at 50C,
the system is sealed and allowed to reach thermal equilibrium.
What is the resulting temperature?
mice = 120 g , cice = 0.5 cal/gC
mAl = 50 g, cAl = 0.22 cal/gC
mw = 100 g, cw = 1 cal/gC
tice = - 10C, tw = 50C
Lf = 80 cal/g
Q LOST = Q GAINED
Q LOST = Q (water) + Q (Al)
Q GAINED = Q (ice) + Q (fusion) + Q (water)
mice = 120 g , cice = 0.5 cal/gC
mAl = 50 g, cAl = 0.22 cal/gC
mw = 100 g, cw = 1 cal/gC
tice = - 10C, tw = 50C
Lf = 80 cal/g
Q LOST = mw cw T + mAl cAl T
100(1)(50- teq ) + 50(0.22)(50- teq) = 5550 - 111 teq
Q GAINED = mice cice T + mice Lf + mw cw T
12(0.5)(0 - (-10)) + 12(80) + 12(1) (teq - 0) = 1020 + 12 teq
5550 - 111 teq = 1020 + 12 teq
teq = 36.8C
14.6 If 10 g of steam at 100C is introduced into a mixture of
200 g of water and 120 g of ice, find the final temperature and
composition of the mixture.
mice = 120 g, cice = 0.5 cal/gC
mw = 200 g, cw = 1 cal/gC
msteaml = 10 g, tsteam = 100C
Lf = 80 cal/g, Lv = 540 cal/g
Heat needed to melt the ice:
Q1 = mice Lf
= 120(80) = 9600 cal
Heat given off by steam :
Q2 + Q3 = msteam Lv + mw cw T
= 10(540) + 10(1)(100 - 0) = 6400 cal
mice = 120 g, cice = 0.5 cal/gC
mw = 200 g, cw = 1 cal/gC
msteaml = 10 g, tsteam = 100C
Lf = 80 cal/g, Lv = 540 cal/g
Heat available = 9600 - 6400 = 3200 cal
mice Lf = 3200 cal
mice = 3200/80 = 40 g
mw = initial mass + melted ice + condensed steam
mw = 200 + (120 - 40) + 10 = 290 g
Final temperature must be 0C