Cutnell & Johnson, 7 th edition
Chapter 12: Problems 20, 34, 97
Chapter 13: Problems 9, 29, 44
*20 As the drawing shows, two thin strips of metal are bolted together at one end and have the same temperature. One is steel, and the other is aluminum. The steel strip is 0.10% longer than the aluminum strip. By how much should the temperature of the strips be increased, so that the strips have the same length?
*34 A solid aluminum sphere has a radius of 0.50 m and a temperature of 75 °C. The sphere is then completely immersed in a pool of water whose temperature is 25 °C. The sphere cools, while the water temperature remains nearly at 25 °C, because the pool is very large. The sphere is weighed in the water immediately after being submerged (before it begins to cool) and then again after cooling to 25 °C. (a) Which weight is larger? Why? (b) Use Archimedes’ principle to find the magnitude of the difference between the weights.
97.
Concept Questions An aluminum can is filled to the brim with a liquid. The can and the liquid are heated so their temperatures change by the same amount. (a) In general, what factors determine how the volume of an object changes when it is heated? (b) Aluminum has a smaller coefficient of volume expansion than the liquid does. Which, if either, will expand more, the can or the liquid? Provide a reason for your answer. (c) How is the volume of liquid that spills over related to the changes in the volume of the can and the liquid?
Problem The can’s initial volume at 5 °C is 3.5x10
-4 m 3 . The coefficient of volume expansion for aluminum is 69x10-6 (C˚) -1 (Table 12-1). When the can and the liquid are heated to 78 ˚C,
3.6x10
-6 m 3 of liquid spills over. What is the coefficient of volume expansion of the liquid?
*9 Review Conceptual Example 5 before attempting this problem. To illustrate the effect of ice on the aluminum cooling plate, consider the drawing shown here and the data contained therein. Ignore any limitations due to significant figures.
(a) Calculate the heat per second per square meter that is conducted through the ice–aluminum combination. (b) Calculate the heat per second per square meter that would be conducted through the aluminum if the ice were not present.
Notice how much larger the answer is in (b) as compared to (a).
29 See Multiple-Concept Example 8. A person eats a dessert that contains 260 Calories.
(This “Calorie” unit, with a capital C, is the one used by nutritionists; 1 Calorie = 4186 J. See
Section 12.7.) The skin temperature of a person is 36 °C and that of her environment is 21 °C.
The emissivity of her skin is 0.75 and its surface area is 1.3 m 2 . How much time would it take for her to emit a net radiant energy from her body equal to the energy contained in this dessert?
44 Concept Questions Part ( a ) of the drawing shows a rectangular bar whose dimensions are
. The bar is at the same temperature as the room (not shown) in which it is located. (a) Is the net radiant power emitted by the bar greater than zero, equal to zero, or less

than zero? Provide a reason for your answer. (b) The bar is then cut, lengthwise, into two equal pieces, as shown in part ( b ) of the drawing. The temperature of the bars does not change. Which situation, if either, emits more power into the room, the single bar in part ( a ) or the two bars in part ( b ) of the drawing? Why? (c) Which situation, if either, absorbs more power from the room, the single bar in part ( a ) or the two bars in part ( b ) of the drawing? Justify your reasoning.
Problem (a) What is the ratio of the power absorbed by the two bars in part ( b ) of the drawing to the single bar in part ( a )? (b) Suppose the temperature of the single bar in part ( a ) is 450 K. What would the temperature (in Kelvins) of the room and the two bars in part ( b ) have to be such that the two bars absorb the same power as the single bar in part ( a )?
20 . REASONING The length of either heated strip is L
0
+  L , where L
0
is the initial length and
 L is the amount by which it expands. The expansion  L can be expressed in terms of the coefficient of linear expansion  , the initial length L
0
, and the change in temperature  T , according to Equation 12.2. To find the change in temperature, we will set the length of the heated steel strip equal to the length of the heated aluminum strip and solve the resulting equation for  T .
SOLUTION According to Equation 12.2, the expansion is  L =  L
0
 T . Using this equation we have
L
0, Steel
 
L
Steel 0, Steel
 
L
0, Aluminum
 
L
Aluminum 0, Aluminum

T

L
Steel

L
Aluminum
L
0, Steel

1
 
Steel

T


L
0, Aluminum

1
 
Aluminum

T

We know that the steel strip is 0.10 % longer than the aluminum strip, so that
L
0, Steel
 
1.0010

L
0, Aluminum
. Substituting this result into the equation above, solving for
 T , and taking values for the coefficients of linear expansion for aluminum and steel from
Table 12.1give


1.0010

L
0, Aluminum

1
 
Steel

T


L
0, Aluminum

1
 
Aluminum

T

0.0010
 
1.0010
 
Steel

Aluminum

T

Aluminum
0.0010
 
1.0010
 
Steel
34.
REASONING AND SOLUTION

23 10

6
  
1  
0.0010
 

 
6
  
1



91 C
 a. The apparent weight of the sphere will be larger after
it cools. This is because the sphere shrinks while cooling, displacing less water, and hence decreasing the buoyant force acting on it. b. The weight of the submerged sphere before cooling is
W
0
=  gV
0
  w gV
0
The weight of the submerged sphere after cooling is
W =  gV
0
  w gV
The weight difference is
 W = W  W
0
=   w g  V where  V = V  V
0
. The volume change is  V =  V
0
 T . Now,  W =   w g  V
0
 T . The sphere’s original volume is
V
0
= (4/3)  r
3 = (4/3)(  )(0.50 m) 3 = 0.52 m 3
The coefficient of volumetric expansion  for aluminum is given in Table 12.1, so we have
  W =  (1.0  10 3 kg/m 3 )(9.80 m/s 2 )[69  10 –6 (C°) –1 ](0.52 m 3 )(  5.0  10 1 C°) 
18 N
97.
CONCEPT QUESTIONS a. According to 
V
0

T
(Equation 12.3), the change  V in volume depends on the coefficient of volume expansion  , the initial volume V
0
, and the change in temperature  T . b. The liquid expands more, because its coefficient of volume expansion is larger and the change in volume is directly proportional to that coefficient. c. The volume of liquid that spills over is equal to the change in volume of the liquid minus the change in volume of the can.

SOLUTION The volume of liquid that spills over the can is the difference between the increase in the volume of the liquid and that of the aluminum can:

V
L 0
  
V
A 0

T
Therefore,

L


V
V
0

T
 
A
 
 
4
 
6 m
3 m
3
     

69 10

6
  
1 
2.1 10

4
  
1
9.
REASONING AND SOLUTION Using Equation 13.1,
Q

( kA T t L
, we obtain
F
Q
H
At L
(1)
Before Equation (1) can be applied to the ice-aluminum combination, the temperature T at the interface must be determined. We find the temperature at the interface by noting that the heat conducted through the ice must be equal to the heat conducted through the aluminum:
Q ice
= Q aluminum
. Applying Equation 13.1 to this condition, we have
L
 ice


L
 aluminum
(2) or
          t
0.0050 m 0.0015 m
The factors A and t can be eliminated algebraically. Solving for T gives T = –24.959 °C for the temperature at the interface. a. Applying Equation (1) to the ice leads to

Q
At

 
  ice

0.0050 m

 2
Since heat is not building up in the materials, the rate of heat transfer per unit area is the same throughout the ice-aluminum combination. Thus, this must be the heat per second per square meter that is conducted through the ice-aluminum combination. b. Applying Equation (1) to the aluminum in the absence of any ice gives:



Q
At

Al

0.0015 m

 2
_29.
REASONING According to the discussion in Section 13.3, the net power P net
radiated by the person is P net
 

4 
T
0
4

, where e is the emissivity,

is the Stefan-Boltzmann constant, A is the surface area, and T and T
0
are the temperatures of the person and the environment, respectively. Since power is the change in energy per unit time (see
Equation 6.10b), the time t required for the person to emit the energy Q contained in the dessert is t = Q / P net
.
SOLUTION The time required to emit the energy from the dessert is t

Q
P net



Q
4 
T
0
4

 
260 Calories
 4186 J
The energy is Q , and the Kelvin temperatures are
1 Calorie
T = 36

C + 273 = 309 K and T
0
= 21

C + 273 = 294 K. The time is t

   
8


260 Calories
 2 
K
4
 
 4186 J
1 Calorie
1.3 m
2
 


309 K
 
K
 4

 

44.
CONCEPT QUESTIONS a. The net radiant power emitted by the bar in part ( a ) of the drawing is zero. The reason is that the temperature of the bar is the same as that of the room, and this temperature does not change. Therefore, the bar emits the same power into the room as it absorbs from the room, so the net radiant power emitted by the bar is zero. b. The two bars in part ( b ) of the drawing emit more power. According to Equation 13.2, the radiant power (or energy per unit time) emitted by an object is /
 
, which is directly proportional to its surface area A . The two bars in part ( b ) have a greater total surface area than the single bar in part ( a ). c. The two bars in part ( b ) of the drawing also absorb more power. From the results of
Concept Question b, we know that the two bars emit more power because of their greater surface area. However, since their temperature does not change, the two bars must absorb as much power as they emit. Thus, they absorb more power from the room than the single bar in part ( a ).
SOLUTION a. The power (or energy per unit time) absorbed by the two bars in part ( b ) of the drawing is given by /
 
2
, where A
2
is the total surface area of the two bars: A
2

28 L
2
0
. The power absorbed by the single bar in A is /
 
1
, where A
1
is the total surface area of the single bar: A
1

22 L
2
0
.The ratio of the power P
2
absorbed by the two bars in part ( b ) to the power P
1 absorbed by the single bar in part ( a ) is
P
2
P
1



4
4


28 L
2
0
22 L
2
0

 
1.27
b. If the power absorbed by the two bars in part ( b ) of the drawing is the same as that absorbed by the single bar in part ( a ), then

1 1
 
2 2
Power absorbed Power absorbed by single bar by two bars
Solving for the temperature of the room and the bars in part ( b ) gives
T
2

T
1
4
A
1
A
2
 
450.0 K

4
22 L
2
0 
424 K
28 L
2
0
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