# Physics 41 Chapter 19 & 20 HW Solutions

```Physics 41 Chapter 19 & 20 HW Solutions
1. Is it possible to convert internal energy to mechanical energy? Explain with examples.
2. The pendulum of a certain pendulum clock is made of brass. When the temperature increases, does the period of
the clock increase, decrease or remain the same? Explain.
3. You need to pick up a very hot cooking pot in your kitchen. Which will protect your hands more: dry pads or wet
pads. Explain.
4. Why are icebergs often surrounded by fog? Water vapor saturates cool air, water condenses into clouds.
5. Some folks can walk on hot coals with bare feet and not burn their feet. Explain how they manage this. Magic or
physics? Wet feet. Specific heat of water and feet is high.
6. The power radiated by a distant star is 4.2 x1027 W. The radius of the star, which may be considered a perfect radiator, is 1.06
x 1010 m. Determine the surface temperature of the star.
Solution: This is a problem of thermal radiation. The rate at which energy is being radiated is proportional to
the “fourth power of the absolute temperature. This behavior is known as Stefan’s Law,” From this behavior
we can see that the energy radiated is also proportional to the surface area of the star.
Ρ =σ AeT 4
T =(
Ρ
)
σ Ae
1
4
e is the emissivity of the object; since we make the assumption that the star is a
perfect radiator, we can conclude that it also is a perfect absorber or a black body;
therefore e=1.
Another assumption is that the star is a perfect sphere, and therefore the surface area is:
=
A 4=
π r 2 | r radius
1
Ρ
4.2 x1027 W
4
=
T (=
)
(
4
4π r 2σ e
4π (1.06 x1010 m) 2 (1)(5.6696 x10−8 WK
=
T 2691.3
=
K
[ 2690 K ]
1
)4
m2
)
7. A granite wall has a thickness of 0.61 m and a thermal conductivity of 2.1 W/(m • C°). The temperature on one face of the
wall is 3.2 °C and 20.0 °C on the opposite face. How much heat is transferred in one hour through each square meter of the
granite wall?
Solution: Power describes the amount of energy given off over a period of time. Since the energy being given off
is only in the form of heat, we can conclude that:
Q
P=
∆t
dT
where k is the thermal conductivity of the
The law of thermal conduction can be described as P = kA
dx
1
1min
material. The change in time is: 1hour (
)(
) = 3600 s
60 min 60 s
dT Th − Tc
=
=
The Temperature gradient
is
| l length
dx
l
T −T
Q
20.0C o − 3.2C o
 208209.8 J 
=
∆tk ( h c ) =
(3600 s )(2.1W
)(
)=
o
mC
m 2 

0.61m
A
l
J
s( )
W Co
sW
J
s
Unit analysis: s (
)( =
) =
=
o
2
2
mC
m
m
m
m2
So we have that 208209.8 Joules transferred per meter square.
8. Steam at 100°C is added to ice at 0°C. (a) Find the amount of ice melted and the final temperature when the mass
of steam is 10.0 g and the mass of ice is 50.0 g. (b) What If? Repeat when the mass of steam is 1.00 g and the mass of
ice is 50.0 g.
(a)
(
)(
)
50.0 10−3 kg 3.33 × 105 J kg =
1.67 × 104 J
Q1 = heat to melt all the ice =×
=
Q2
=
( heat to raise temp of ice to 100°C)
C)
( 50.0 × 10−3 kg ) ( 4 186 J kg ⋅°C) (100°=
2.09 × 104 J
Thus, the total heat to melt ice and raise temp to 100°C
= 3.76 × 104 J
(
)(
)
heat available
=×
Q3 =
10.0 10−3 kg 2.26 × 106 J kg =
2.26 × 104 J
as steam condenses
Thus, we see that Q3 > Q1 , but Q3 < Q1 + Q2 .
Therefore, all the ice melts but T f < 100°C . Let us now find T f
Qcold = −Qhot
( 50.0 × 10 kg )( 3.33 × 10 J kg ) + ( 50.0 × 10 kg ) ( 4 186 J kg ⋅°C) (T − 0°C)
= − (10.0 × 10 kg )( −2.26 × 10 J kg ) − (10.0 × 10 kg ) ( 4 186 J kg ⋅°C) ( T − 100°C)
−3
−3
5
f
−3
−3
6
f
From which, =
T f 40.4°C .
(b)
Q1 = heat to melt all ice
= 1.67 × 104 J [See part (a)]
(
)(
)
heat given up
Q2 =
=
10−3 kg 2.26 × 106 J kg =
2.26 × 103 J
as steam condenses
heat given up as condensed
=
Q3
= 10−3 kg ( 4 186
=
J kg ⋅ °C) ( 100°C) 419 J
steam cools to 0°C
(
)
Note that Q2 + Q3 < Q1 . Therefore, the final temperature will be 0°C with some ice remaining. Let
us find the mass of ice which must melt to condense the steam and cool the condensate to 0°C.
mL f = Q2 + Q3 = 2.68 × 103 J
2.68 × 103 J
=
8.04 × 10−3 kg =
8.04 g .
Thus, m =
3.33 × 105 J kg
Therefore, there is 42.0 g of ice left over .
9. How much energy is required to change a 40.0-g ice cube from ice at –10.0°C to steam at 110°C?
The heat needed is the sum of the following terms:
=
Qneeded
( heat to reach melting point ) + ( heat to melt )
+ ( heat to reach melting point ) + ( heat to vaporize) + ( heat to reach 110°C)
Thus, we have
(
=
Qneeded 0.040 0 kg ( 2 090 J kg ⋅ °C) ( 10.0°C) + 3.33 × 105 J kg

(
)
)
+ ( 4 186 J kg ⋅ °C) ( 100°C) + 2.26 × 106 J kg + ( 2 010 J kg ⋅ °C) ( 10.0°C) 

=
Qneeded
1.22 × 105 J
10. A 50.0-g copper calorimeter contains 250 g of water at 20.0°C. How much steam must be condensed into the
water if the final temperature of the system is to reach 50.0°C?
Qcold = −Qhot
( mw cw + mccc ) (T f − Ti ) =−ms  − Lv + cw (T f − 100) 
0.250 kg ( 4 186 J kg ⋅ °C) + 0.050 0 kg ( 387 J kg ⋅ °C)  ( 50.0°C − 20.0°C)
= − ms  −2.26 × 106 J kg + ( 4 186 J kg ⋅ °C) ( 50.0°C − 100°C) 
=
ms
3.20 × 104 J
= 0.012
=
9 kg
2.47 × 106 J kg
12.9 g steam
11. Following a collision in outer space, a copper disk at 850°C is rotating about its axis with an angular
speed of 25.0 rad/s. As the disk radiates infrared light, its temperature falls to 20.0°C. No external torque
acts on the disk.
(a) Does the angular speed change as the disk cools off? Explain why.
(b) What is its angular speed at the lower temperature?
(c) Find the change in kinetic energy of the disk.
(d) Find the change in internal energy of the disk.
(e) Find the amount of energy it radiates.
The initial moment of inertia of the disk is
(a)
No torque acts on the disk so its angular momentum is constant. Its moment of inertia decreases as
it contracts so its angular speed must increase .
(b)
I i ω=
I f ω=
i
f
1
1
1
1
2
2
M Ri2ω=
M R2f ω=
M Ri + Ri α∆T  ω=
M Ri2 1 − α ∆T  ω f
i
f
f
2
2
2
2
−2
ω=
ω i 1 − α ∆T  =
f
( (
25.0 rad s
)
1 − 17 × 10−6 1 C° 830°C
)
25.0 rad s
=
0.972
=
2
25.7 rad s
(c)
(
)
1
1
1
1
4
M R2 = ρVR2 = ρπ R2tR2 = 8 920 kg m 3 π ( 28 m ) 1.2 m =
1.033 × 1010 kg ⋅ m 2
2
2
2
2
The rotation speeds up as the disk cools off, according to
I iωi = I f ω f
1
1
1
2
2
M=
Ri2ω i
MR
=
M Ri2 ( 1 − α ∆T ) ω f
fω f
2
2
2
1
1
25 rad s
25.720 7 rad s
=
ω f ω=
=
i
2
2
1 − 17 × 10−6 1 °C 830°C 
(1 − α ∆T )


(
)
The kinetic energy increases by
(
)
1
1
1
1
1
I f ω 2f − I i ω i2=
I i ω i ω f − I i ω i2=
I iωi ω f − ωi
2
2
2
2
2
1
= 1.033 × 1010 kg ⋅ m 2 ( 25 rad s) 0.720 7 rad s =9.31 × 1010 J
2
(d)
∆Eint = mc∆T = 2.64 × 107 kg ( 387 J kg ⋅°C) ( 20°C − 850°C) =
(e)
As 8.47 × 1012 J leaves the fund of internal energy, 9.31 × 1010 J changes into extra kinetic energy,
and the rest, 8.38 × 1012 J is radiated.
−8.47 × 1012 J
```