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Chapter 11 Solutions









ANSWERS TO MULTIPLE CHOICE QUESTIONS



1. From the mechanical equivalent of heat, 1 cal  4.186 J . Therefore,



 4.186 J 



3.50  103 cal  3.50  103 cal 

 1 cal 



 1.47  10 4 J





and (b) is the correct choice for this question.





 1 cal   1 Cal 

2. 

7.80  105 J  7.80  10 5 J   

 4.186 J   103 cal 



 186 Cal , so (a) is the correct



choice.





3. The required energy input is



Q  mc  T    5.00 kg  128 J kg  °C   327°C  20.0°C   1.96  10 5 J



and the correct response is (e).



4. 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  10 5 J kg  6.66  10 5 J



Once this is done, there is Q2  Qtotal  Q1  9.30  105 J  6.66  105 J  2.64  105 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).



5. 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







 J   25°C  14°C 

P   0.10

s  m  °C 

 48.0 m 2  4.00  10 2 m  1.3  103 J s  1.3  103 W

   

and the (d) is the correct answer.

6. 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  10 8 W m2  K 4 is a constant. Thus, for the given spherical object



A  

4 r 2 , we have





P   5.669 6  10 8 W m2  K 4  4  2.00 m   0.450    408 K    298 K  

2 4 4



 



yielding P  2.54  104 W , so (e) is the correct choice.



7. 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





Q7.50  105 J  1 min 

t    750 s   12.5 min

P 1.00  10 3 J s  60 s 

making (d) the correct choice.

8.    

We use Qhot  Qcold or mx cx T f  Tx,i  mw cw Tf  Tw,i to compute the specific heat

of the unknown material and find









cx 



mw cw T f  Tw, i    0.400 kg   4 186 J kg  °C   36.0°C  20.0°C   1.82  10 3 J kg  °C

 mx T f  Tx, i   0.250 kg   36.0°C  95.0°C 





which is a match for the specific heat of Beryllium, so (b) is the correct choice.



9. 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. With eA  eB , rA  2rB , and TA  2TB , the ratio of the power output of A to that of B is





2 4

PA  AA eA TA

4

4 rA TA

2 4 r  T 

  A   A    2   2    2   64

2 4 6

 

PB  AB eB TB

4 4 rB B

2T 4

 rB   TB 





making (e) the correct choice.

11. 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. 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).



ANSWERS TO EVEN-NUMBERED CONCEPTUAL QUESTIONS

2. In winter the produce is protected from freezing. The specific heat of Earth is so high that soil

freezes only to a depth of a few inches in temperate regions. Throughout the year the temperature

will stay nearly constant day and night. Factors to be considered are the insulating properties of

the soil, the absence of a path for energy to be radiated away from or to the vegetables, and the

hindrance of the formation of convection currents in the small, enclosed space.



4. The high thermal capacity of the barrel of water and its high heat of fusion mean that a large

amount of energy would have to leak out of the cellar before the water and produce froze solid.

Evaporation of the water keeps the relative humidity high to protect foodstuffs from drying out.



6. Yes, if you know the specific heat of zinc and copper, you can determine the relative fraction of

each by heating a known weight of pennies to a specific initial temperature, say 100° C, then

dump them into a known quantity of water, at say 20° C. The equation for conservation of energy

will be



mpennies  x  cCu   1  x  c Zn   100°C  T   mwater cwater  T  20°C 

 

The equilibrium temperature, T, and the masses will be measured. The specific heats are known,

so the fraction of metal that is copper, x, can be computed.



8. Write mwater cwater  1°C    air V  cair  1°C  , to find





V 

mwater cwater



 

1.0  103 kg  4 186 J kg  °C 

 3.2  103 m3

air cair  

1.3 kg m3 1.0  103 J kg  °C 

10. The black car absorbs more of the incoming energy from the Sun than does the white car, making

it more likely to cook the egg.



12. Keep them dry. The air pockets in the pad conduct energy slowly. Wet pads absorb some energy

in warming up themselves, but the pot would still be hot and the water would quickly conduct a

lot of energy to your hand.



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