PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HOMEWORK PROBLEMS Chapter 11:
ENERGY IN THERMAL PROCESSES
HW-11
PART-A: Hand in your answers in class on scantron on Monday 29-November-2010. Write your
name, class (1401) and HW # 11 on the scantron.
1. Convert 3.50 x 103 cal to an equivalent number of joules.
(a) 2.74 X 104 J
(b) 1.47 X 104 J
(c) 3.24 X 104 J
(d) 5.33 X 104 J
From the mechanical equivalent of heat, 1 cal  4.186 J . Therefore,
 4.186 J 
3.50  103 cal  3.50  10 3 cal 
 1.47  10 4 J
 1 cal 


2. Convert 7.80 x 105 J to the equivalent number of Calories (1 Cal = 1 000 cal).
(a) 186 Cal
(b) 238 Cal
(c) 325 Cal
(d) 418 Cal (e) 522 Cal
 1 cal   1 Cal 
7.80  105 J  7.80  105 J 
 186 Cal , so (a) is the correct choice. (note: cal vs Cal)

 4.186 J   103 cal 


3. How much energy is required to raise the temperature of 5.00 kg of lead from 20.0°C to its
melting point of 327°C?
(a) 4.04 X 105 J
(b) 1.07 X 105 J
(c) 8.15 X 104 J
(d) 2.13 X 104 J
5
(e) 1.96 X 10 J
The required energy input is
Q  mc  T    5.00 kg  128 J kg  °C   327°C  20.0°C   1.96  105 J
and the correct response is (e).
4. If 9.30 X 105 J of energy are transferred to 2.00 kg of ice at 0°C, what is the final temperature
of the system?
(a) 22.4°C
(b) 14.2°C
(c) 31.5°C
(d) 18.0°C
(e) 0°C
The energy which must be added to the 0°C ice to melt it, leaving liquid at 0°C, is
Q1  mL f   2.00 kg  3.33  105 J kg  6.66  105 J


Once this is done, there is Q2  Qtotal  Q1  9.30  105 J  6.66  105 J  2.64  10 5 J of energy
still available to raise the temperature of the liquid. The change in temperature this produces is
Q2
2.64  105 J
T  T f  0°C 

 31.5°C
mcwater
 2.00 kg   4 186 J kg  °C 
so the final temperature is T f  0°C  31.5°C  31.5°C and the correct choice is (c).
1
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-11
5. A wall made of wood 4.00 cm thick has area of 48.0 m2. If the temperature inside is 25°C and
the temperature outside is 14°C, at what rate is thermal energy transported through the wall by
conduction?
(a) 82 W
(b) 210 W
(c) 690 W
(d) 1.3 X 103 W
(e) 2.1 X 103 W
The rate of energy transfer by conduction through a wall of area A and thickness L is
P  kA  Th  Tc  L , where k
is the thermal conductivity of the material making up the wall, while Th and Tc are the temperatures on
the hotter and cooler sides of the wall, respectively. For the case given, the transfer rate will be
 25°C  14°C   1.3  103 J s  1.3  10 3 W
J


P   0.10
48.0 m 2


s  m  °C 
4.00  10 2 m




and the (d) is the correct answer.
6. A granite ball of radius 2.00 m and emissivity 0.450 is heated to 135°C, whereas the ambient
temperature is 25.0°C. What is the net power radiated from the ball?
(a) 425 W
(b) 3.55 X 104 W
(c) 145 W
(d) 2.01 X 103 W
(e) 2.54 X 104 W
The power radiated by an object with emissivity e, surface area A, and absolute temperature T,
in a location with absolute ambient temperature T0 , is given by P   Ae T 4  T04

where   5.669 6 
A 
10 8
W
 , we have
  5.669 6  10 8
m2

K4

is a constant. Thus, for the given spherical object
4 r 2
P
yielding

W m 2  K 4 4  2.00 m 
2
 0.450   408 K 4   298 K 4 
P  2.54  104 W , so (e) is the correct choice.
7. How long would it take a 1.00 X 103 W heating element to melt 2.00 kg of ice at −20.0°C,
assuming all the energy is absorbed by the ice?
(a) 4.19 s
(b) 419 s
(c) 555 min (d) 12.5 min
(e) 2.00 h
The temperature of the ice must be raised to the melting point, T  20.0°C , before it will start to
melt. The total energy input required to melt the 2.00-kg of ice is
Q  mcice  T   mL f   2.00 kg    2 090 J kg  °C   20.0°C   3.33  10 5 J kg   7.50  10 5 J
The time the heating element will need to supply this quantity of energy is
Q
7.50  105 J
 1 min 
t 

 750 s 
 12.5 min
3
 60 s 
P 1.00  10 J s
making (d) the correct choice.
8. An unknown element of mass 0.250 kg, initially at 95.0°C, is dropped into 0.400 kg of water
at 20.0°C contained in an insulated cup of negligible mass and specific heat. If the equilibrium
temperature is 36.0°C, which of the following is the most likely identity of the substance?
(a) aluminum
(b) beryllium
(c) cadmium (d) iron
(e) gold



We use Qhot  Qcold or mx cx T f  Tx,i  mw cw T f  Tw,i
unknown material and find
2
 to compute the specific heat of the
PHYS-1401: College Physics-I
cx 

T
CRN 55178
Khalid Bukhari
   0.400 kg   4 186 J kg  °C   36.0°C  20.0°C   1.82  10
 0.250 kg   36.0°C  95.0°C 

mw cw T f  Tw, i
 mx
f
 Tx, i
HW-11
3
J kg  °C
which is a match for the specific heat of Beryllium, so (b) is the correct choice.
9. An amount of energy is added to ice, raising its temperature from −10°C to −5°C. A larger
amount of energy is added to the same mass of water, raising its temperature from 15°C to 20°C.
From these results, what can we conclude?
(a) Overcoming the latent heat of fusion of ice requires an input of energy.
(b) The latent heat of fusion of ice delivers some energy to the system.
(c) The specific heat of ice is less than that of water.
(d) The specific heat of ice is greater than that of water.
(e) More information is needed to draw any conclusion.
Since less energy was required to produce a 5°C rise in the temperature of the ice than was required to
produce a 5°C rise in temperature of an equal mass of water, we conclude that the specific heat of ice
c  Q m  T  is less than that of water. Thus, choice (d) is correct.


10. blank
11. A person shakes a sealed, insulated bottle containing coffee for a few minutes. What is the
change in the temperature of the coffee?
(a) a large decrease (b) a slight decrease (c) no change
(d) a slight increase
(e) a large increase
By agitating the coffee inside this sealed, insulated container, the person is raising the internal energy of
the coffee, which will result is a rise in the temperature of the coffee. However, doing this for only a few
minutes, the temperature rise will be quite small. The correct response to this question is (d).
12. A poker is a stiff, nonflammable rod used to push burning logs around in a fireplace. Suppose
it is to be made of a single material. For best functionality and safety, should the poker be made
from a material with
(a) high specific heat and high thermal conductivity,
(b) low specific heat and low thermal conductivity,
(c) low specific heat and high thermal conductivity,
(d) high specific heat and low thermal conductivity, or
(e) low specific heat and low density?
One would like the poker to be capable of absorbing a large amount of energy, but undergo a small rise in
temperature. This means it should be made of a material with a high specific heat capacity. Also, it is
desirable that energy absorbed by the end of the poker in the fire be conducted to the person holding the
other end very slowly. Thus, the material should have a low thermal conductivity. The correct choice is (d).
13. Lead pellets, each of mass 1.00 g, are heated to 200°C. How many pellets must be added to
500 g of water that is initially at 20.0°C to make the equilibrium temperature 25.0°C?
Neglect any energy transfer to or from the container.
a) 367
b) 467
c) 552
d) 321
3
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-11
If N pellets are used, the mass of the lead is Nmpellet . Since the energy lost by the lead must equal the
energy absorbed by the water,
Nmpellet c  T 
  mc  T  
water
lead
N 

or the number of pellets required is
mw cw  T  w
mpellet clead T
lead
 0.500 kg   4 186 J kg  °C   25.0°C  20.0°C 
1.00  10-3 kg  128 J kg  °C   200°C  25.0°C 
 467
14. An aluminum cup contains 225 g of water and a 40-g copper stirrer, all at 27°C. A 400-g
sample of silver at an initial temperature of 87°C is placed in the water. The stirrer is used to
stir the mixture until it reaches its final equilibrium temperature of 32°C. Calculate the mass
of the aluminum cup.
a) 80 g
b) 100 g
c) 120 g
d) 210 g
The total energy absorbed by the cup, stirrer, and water equals the energy given up by the silver sample.
Thus,
 mc cAl  ms cCu  mw cw   T  w   mc T 
Ag
Solving for the mass of the cup gives

T Ag
1 
 mAg cAg
mc 
 ms cCu  mw cw  ,
cAl 
 T  w



or
 87  32   40 g 387  225 g 4 186   80 g
1 
mc 
   


  400 g   234 
900 
 32  27 



15. A 50-g ice cube at 0°C is heated until 45 g has become water at 100°C and 5.0 g has
become steam at 100°C. How much energy was added to accomplish the transformation?
a) 49 J
b) 58 J
c) 21 J
d) 31 J
The total energy input required is
Q   energy to melt 50 g of ice 
  energy to warm 50 g of water to 100°C 
+  energy to vaporize 5.0 g water 
  50 g  L f   50 g  cwater  100°C  0°C    5.0 g  Lv
Thus,

J 
Q   0.050 kg   3.33  10 5
kg 


J 
  0.050 kg   4 186
100°C  0°C 
kg  °C 


J 
 5.0  10 3 kg  2 .26  10 6
kg 


which gives Q  4.9  10 4 J  49 kJ .
4

PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-11
16. A 100-g cube of ice at 0°C is dropped into 1.0 kg of water that was originally at 80°C.
What is the final temperature of the water after the ice has melted?
a) 65 C
b) 74 C
c) 91 C
d) 103 C
The conservation of energy equation for this process is
 energy to melt ice    energy to warm melted ice to T    energy to cool water to T 
Or
mice L f  mice cw  T  0°C   mw cw  80°C  T 
This yields
So
17.
T 
mw cw  80°C   mice L f
 mice
1.0 kg   4 186
T 
 mw  c w

J kg  °C   80°C    0.100 kg  3.33  10 5 J kg
1.1 kg   4 186
J kg  °C 

 65 °C
How much energy is required to change a 40-g ice cube from ice at −10°C to steam at
110°C?
a) 0.08 MJ
b) 1.0 KJ
c) 0.12 MJ
d) 0.22 MJ
The energy required is the following sum of terms:
Q   energy to reach melting point 
  energy to melt    energy to reach boiling point 
  energy to vaporize    energy to reach 110°C 
Mathematically,
Q  m  cice  0°C   10°C    L f  cw 100°C  0°C   Lv  csteam 110°C  100°C  
This yields


J 
J 
Q  40  10 3 kg   2 090
10°C    3.33  10 5


kg  °C 
kg 





J 
J  
J 
  4 186
100°C    2.26  106
  2 010
10°C  





kg  °C 
kg  
kg  °C 





or
Q  1.2  105 J  0.12 MJ
18.
A glass windowpane in a home is 0.62 cm thick and has dimensions of 1.0 m X 2.0 m.
On a certain day, the indoor temperature is 25°C and the outdoor temperature is 0°C. What
is the rate at which energy is transferred by heat through the glass
a) 6.1 kW
b) 6.8 kW
c) 13.2 kW
d) 3.4 kW
The rate of energy transfer by conduction through a material of area A, thickness L, with thermal
conductivity k, and temperatures Th  Tc on opposite sides is P  k A  Th  Tc  L . For the given
windowpane, this is
 25°C  0°C   6.8  103 J s  6.8  103 W
J


 1.0 m   2.0 m  
P   0.84


s  m  °C 
0.62  10 2 m
5