II. Heat
1. Heat and energy
A little bit of history:
• Old theories
• Robert von Mayer (1814-1878)
• James Joule (1818-1889)
• Hermann von Helmholtz (1821-1894)
1 cal = 4.186 J
Calorie: the amount of heat required to raise the temperature
of 1 gram of water from 14.5 ◦C to 15.5 ◦C.
2. Specific heat
dQ  cmdT
Q  cmT


1 dQ
c
m dT
cwater  1cal / g  C   4190 J / kg  K 
1 cal = 4.186 J
2a. Heat capacity
Q  C T
C  cm
dQ  CdT
dQ
C
dT
2. Calorimetry
 Q  0
ΔQ=cmΔT
i
Example: 200 g of water at the temperature of 80°C is placed into a
50 g glass at 20°C. What is the final temperature of the system.
mw = 200g
Tw = 80º C
cw=1.0cal/(gCº)
mg = 50g
Tg = 20º C
cg = 0.20cal/(gCº)
T=Tf - ?
 Qi  0
Qw  Qg  0
Qw  cwmw (T  Tw )  0
Qg  c g mg (T  Tg )  0
cw mw (T  Tw )  cg mg (T  Tg )  0
T
cw mwTw  c g mg Tg
c w mw  c g m g
 ci mi (T  T )i  0
c mT

T
c m
i
i i
i
i
3. Phase transitions.
Q = ±mL
ΔQ = ±ΔmL
Latent heat: energy required for a
material to change phase, even though
its temperature is not changing.
Example: 100 g of ice at the temperature of -40°C should be melted
and heated to the temperature of 20°C. Calculate needed heat.
m= 100g
Ti = -40º C
ci=0.5cal/(gCº)
T = 20º C
cw =1.0cal/(gCº)
Tc = 0º C
L = 80 cal/g
ΔQ - ?
Q  c i m(Tc  Ti )  mL  c w m(T  Tc ) 
 mc i (Tc  Ti )  L  c w (T  Tc )
Q  100 g * [0.5cal/(gC º ) * (0º C  (40º C)) 
 80cal / g 
 1.0cal/(gC º ) * (20º C  0º C)] 
 100 g * [20  80  20]cal / g  12000cal
Example: Ice and Water
200 g of ice at -40º C is mixed with 300 g of water at 20º C.
mw = 300g
What will happen with ice and water?
mi = 200g
Tw = 20º C
Ti = -40º C
ci=0.5cal/(gCº)
cw =1.0cal/(gCº)
Tc = 0º C
L = 80 cal/g




Qw  cw mw (Tc  Tw )  1.0cal / gC   300 g 0 C  20 C  6000cal





Qi  ci m i (Tc  Ti )  0.5cal / gC   200 g 0 C   40 C  4000cal
Qiw  mL  m  80 cal/g
Qw  Qi  Qiw  6000cal  4000cal  m  80 cal/g  0
m  25g
T  0 C
25 g of ice will be melted
The temperature of the mixture will be 0º C
Example
Energy is required for a material
to change phase, even though
its temperature is not changing.
Avogadro number and number of moles
One mole is the amount of substance that contains as many
elementary entities as there are atoms in 12g of carbon -12
n
N A  6.022  1023 molecules / mol
N
NA
n - number of moles
N - number of molecules
NA - Avogadro number
msample - total mass of sample
m - mass of one molecule
μ=M - molar (atomic) mass (“weight”)
Molecular weight
msample  mN  mnN A  n
mN A  
Molar specific heat (molar heat capacity)
Q  cM nT
c M n  cm sample
dQ  cM ndT
cM  c
cM
1 dQ

n dT
Question
A beaker contains 0.200 kg of
water. The heat capacity of the
water is ___ J/˚C.
Assume: the specific heat of
water is 4000 J/(kg·˚C).
1. 200
2.
3.
4.
400
600
800
Question
The heat capacity of the water in
a massless beaker is 800 J/˚C.
The amount of heat needed to
increase the temperature of the
water from 15.0 ˚C to 30.0 ˚C is
___ kJ.
1. 4
2. 8
3. 12
4. 16
Question
A liter of water is put in a sauce pan
on a stove. In order to heat the water
at 10 ˚C/minute requires a heating
power of ___ Watts.
(Ignore the heat capacity of the pan.
Assume the heat capacity of a liter of
water is 4200 J/˚C.)
1. 500
2. 700
3. 900
4. 1100
Question
Block A: 1.0 kg, c = 400 J/(kg·˚C), initial temperature = 100 ˚C
Block B: 2.0 kg, c = 200 J/(kg·˚C), initial temperature = 0 ˚C
After the blocks are put into thermal
contact and reach equilibrium, the
final temperature is __ ˚C.
1.
2.
3.
4.
50
30
70
40
Example
How much heat does it take to heat 1.00 kg of water ice from
10 ˚C up to 100 ˚C and boil the water away as steam?
Given: cice  2.10 kJ/kg·°C
cwater  4.19 kJ/kg·°C
Lf  334 kJ/kg Lv  2256 kJ/kg
1. Heat the ice up to 0 °C
Q1  mcice T  (1 kg)(2.1 kJ/kg·°C)(10 °C)
= 21.0 kJ.
2. Melt the ice at 0 °C
Q2  mLf  (1 kg)(334 kJ/kg) = 334 kJ
3. Heat the liquid water up to 100 °C
Q3  mwater cwater T
 (1 kg)(4.19 kJ/kg·°C)(100 °C)
= 419 kJ
4. Boil off the water into steam at 100 °C
Q4  mLv  (1 kg)(2256 kJ/kg) = 2256 kJ
Total
Q  Q1  Q2  Q3  Q4
= 21 kJ + 334 kJ + 419 kJ + 2256 kJ
= 774 kJ + 2256 kJ
Q = 3030 kJ = 3.03 MJ
Most (~ 3/4) of the heat is used in boiling
off the water.
Example
A 10 g ice cube at 0 ˚C is put into 100 g of water at 20 ˚C. Does the ice
cube melt completely? If not, the final temperature of the water + ice cube
would be 0 ˚C. If so, what is the final temperature of the water?
Lf  334 kJ/kg
cwater  4.19 kJ/kg·°C
The heat needed to be added to the ice to
melt it is
Qice  mice Lf  (0.010 kg)(334 kJ/kg) = 3.34 kJ.
The heat given off by the liquid water when it
is cooled to 0 °C is
| Qwater |  mwater c | T |
 (0.100 kg)(4.19 kJ/kg·°C)(20 °C)
= 8.38 kJ.
Thus there is more than enough thermal
energy in the water to completely melt the ice.
The heat absorbed by the ice to raise its
temperature to the final temperature T2 is
Qice  mice Lf  mice cwater (T2  Tice 1 )
 mice Lf  mice cwaterT2 .
The heat absorbed by the water to decrease
its temperature to the final temperature is
Qwater  mwater cwater (T2  Twater 1 ).
The sum of these two heats adds to zero:
mice Lf  mice cwaterT2  mwater cwater (T2  Twater 1 )  0.
Solving for T2 gives
m c T
 mice Lf
T2  water water water 1
(mice  mwater )cwater
(0.100 kg)(4.19 kJ/kg·°C)(20 °C)  3.34 kJ
=
(0.010 kg + 0.100 kg)(4.19 kJ/kg·°C)
= 10.9 °C .