Homework Chapter 20 Solutions!
20.7
20.8
20.13
20.17
20.30
20.35
20.39
20.44
20.61
20.69
page 1
Problem 20.7
In cold climates including the northern United States a house can be built with very large windows
facing south to take advantage of heating. Sunlight shining in during the daytime is absorbed by
the floor Interior walls and objects in the room raising the temperature to 38 °C. If the house is
well insulated you may model it as losing energy by heat steadily at the rate 6000 W on a day in
April when the average exterior temperature is 4 °C and when the conventional heating system is
not used at all. During the period between 5 pm and 7 am the temperature of the house drops and
a sufficiently large thermal mass is required to keep it from dropping too far. The thermal mass
can be a large quantity of stone in the floor and the interior walls exposed to sunlight. Use a
specific heat of 850 J/kg•K for the stone.
What's mass of stone is required if the temperature is not to drop below 18 °C overnight?
Solution
The amount of radiation heat loss is
⎛ 3600 s ⎞⎟
⎟ = 3.0240 ×108 J
Qloss = 6000 W ⋅ 14 hours ⎜⎜
⎜⎝ 1 hour ⎟⎟⎠
This heat comes from the temperature change of the stone.
Qstone = −3.0240 ×108 J = mstonecstone (Tf −Ti ) = mstone (850
−3.0240 ×108 J = mstone (850
J )(18
kg⋅K
J )(18
kg⋅K
°C − 38 °C )
°C − 38 °C ) ⇒ mstone = 1.7788 ×104 kg
mstone = 1.78 ×104 kg
page 2
Problem 20.8
A aluminum cup of mass 200 g contains 800 g of water in thermal equilibrium at 80 °C. The
combination of cup and water is cooled uniformly so that the temperature decreases by 1.5 °C per
minute.
At what rate is energy being removed by heat?
Solution
The heat flow out of the system every minute is
Qcup + Qwater = mcupcal ΔT + mwcw ΔT = (0.2 kg)(900
J )(−1.5
kg⋅K
K) + (0.8 kg)(4186
J )(−1.5
kg⋅K
K)
Qcup + Qwater = 5293 J
The heat flow rate is
Qcup + Qwater
Δt
=
5293 J
= 88.2 W
60 s
page 3
Problem 20.13
Two thermally insulated vessels are connected by a valve. One vessel of volume 16.8 L contains
oxygen at a temperature of 300 K and a pressure of 1.75 atm. The other vessel of volume 22.4 L
contains oxygen at a temperature of 450 K and a pressure of 2.25 atm. When the valve is
opened, the gases mix and come to equilibrium.
What is the final temperature?
What is the final pressure?
Solution
The number of moles of oxygen in vessel 1 is
PV = nRT
⇒ n=
Pa )(16.8 ×10−3 m 3 )
(1.75 atm)(101, 325 atm
PV
=
= 1.1944 mol
J )(300 K)
RT
(8.314 mol⋅K
The number of moles of oxygen in vessel 2 is
PV = nRT
⇒ n=
Pa )(22.4 ×10−3 m 3 )
(2.25 atm)(101, 325 atm
PV
=
= 1.3650 mol
J )(450 K)
RT
(8.314 mol⋅K
The mass of oxygen in vessel 1 is
⎛ 32 g ⎞⎟
⎟ = 38.221 g
m = 1.1944 mol ⎜⎜
⎜⎝ 1 mol ⎟⎟⎠
The mass of oxygen in vessel 2 is
⎛ 32 g ⎞⎟
⎟ = 43.680 g
m = 1.3650 mol ⎜⎜
⎜⎝ 1 mol ⎟⎟⎠
The heat flow between the oxygen in the two vessels add to zero.
Q1 + Q2 = 0
m1coxy ΔT1 + m2coxy ΔT2 = 0
(38.221)(Tf − 300) + (43.680)(Tf − 450) = 0
Tf = 380 K
The final pressure is
PV = nRT
⇒ P=
J )(380 K)
(2.5594 mol)(8.314 mol⋅K
nRT
=
= 2.0627 ×105 Pa = 2.04 atm
−3
3
V
39.2 ×10 m
page 4
Problem 20.17
Steam at 100 °C is added to ice at 0 °C. Find the amount of ice melted and the final temperature
when the mass of the steam is 10 g and the mass of ice is 50 g. Repeat for 1 g of steam and 50 g
of ice.
Solution
Here is the summary of the results.
T (0 °C)
ice (0.05 kg)
T (100 °C)
steam (0.01 kg)
melt (0 °C)
16,650 J
condense (100 °C)
22,600 J
T (0 °C)
water (0.05 kg)
T (100 °C)
T (100 °C)
20,930 J
condense (100 °C)
T (28.128 °C)
water (0.05 kg)
T (100 °C)
water (0.01 kg)
T (40.357 °C)
water (0.05 kg)
T (40.357 °C)
water (0.01 kg)
water
(0.0073673 kg)
steam
(0.0026327 kg)
5,950 J
For all of the ice to melt, the heat required is
Q = Lf ∆m = (3.33 ×105
J )(0.05
kg
kg) = 16, 650 J
For all of the steam to condense, the heat required is
Q = Lv ∆m = (2.26 ×106
J )(0.01
kg
kg) = 22, 600 J
All of the ice melt. The amount of steam that condense is
(2.26 ×106
J )Δm
kg
= 16, 650 J ⇒ Δm = 0.0073673 kg
The remaining steam is
m = 0.01 kg − 0.0073673 kg = 0.0026327 kg
Next, the water can all heat up to 100 °C. This requires too much heat.
Q = mc ∆T = (0.05 kg)(4186
J )(100
kg⋅°C
°C ) = 20, 930 J
So all of the steam go to water. The cold water warms to
(0.05 kg)(4186
J )(T
f
kg⋅°C
− 0 °C ) = 5, 950 J ⇒ Tf = 28.128 °C
The last step is mixing water.
(0.05 kg)(Tf − 28.128 °C ) + (0.01 kg)(Tf − 100 °C ) = 0
0.06Tf − 1.4214 − 1 = 0
⇒ Tf = 40.357 °C
The final temperature is 40.4 °C.
page 5
For 1 g of steam,
T (0 °C)
ice (0.05 kg)
T (100 °C)
steam (0.001 kg)
16,650 J
melt (0 °C)
water
(0.0067868 kg)
T (0 °C)
melt (0 °C)
water
(0.0080439 kg)
T (0 °C)
condense (100 °C)
2,260 J
ice
(0.043213 kg)
T (100 °C)
water (0.001 kg)
14,390 J
T (0 °C)
418.6 J
ice
(0.041956 kg)
T (0 °C)
water (0.001 kg)
This time around, all of the steam condenses first. The ice that melts is
(3.33 ×105
J )Δm
kg
= 2, 260 J ⇒
∆m = 0.0067868 kg
The next possible steps are melting of all of the ice using 14,390 J of heat and the cooling of all of
the hot water to 0 °C using
Q = mc ∆T = (0.001 kg)(4186
J )(100
kg⋅°C
°C ) = 418.6 J
This means all of the hot water cools to 0°C. The amount of ice that melts here is
(3.33 ×105
J )Δm
kg
= 418.6 J ⇒
∆m = 0.0012571 kg
The total amount of ice that melts is 0.0080439 g.
page 6
Problem 20.30
Solution
Put the numbers in a table again starting with the givens.
∆Eint
AB
BC
CD
DA
Q
W
100 kJ
–150 kJ
The change in the internal energy is zero during CD. The heat is zero during AB.
∆Eint
AB
BC
CD
DA
Q
0J
100 kJ
W
0J
–150 kJ
The work done during BC is
W = −(3)(101, 325 Pa)(0.31 m 3 ) = −94.232 kJ
The work done during DA is
W = −(1)(101, 325 Pa)(−1 m 3 ) = 101.325 kJ
The work done during CD is
⎛V ⎞
⎛V ⎞
⎛ 1.2 ⎞⎟
⎟ = −133.58 kJ
W = −nRTCD ⋅ ln ⎜⎜ D ⎟⎟⎟ = −PCVC ⋅ ln ⎜⎜⎜ D ⎟⎟⎟ = −(3)(101, 325)(0.4) ⋅ ln ⎜⎜
⎜⎝ 0.4 ⎟⎟⎠
⎜⎝VC ⎟⎠
⎝VC ⎟⎠
page 7
∆Eint
AB
BC
CD
DA
Q
0J
100 kJ
0J
–150 kJ
W
–94.23 kJ
–133.58
101.33 kJ
This means the heat during CD is +133.58 kJ making total heat through the entire cycle 83.58 kJ.
∆Eint
AB
BC
CD
DA
0J
Q
0J
100 kJ
133.58 kJ
–150 kJ
W
–94.23 kJ
–133.58
101.33 kJ
This means the total work through the entire cycle is –83.58 kJ. So the work during AB must be
42.9 kJ.
∆Eint
AB
BC
CD
DA
0J
Q
0J
100 kJ
133.58 kJ
–150 kJ
W
42.9 kJ
–94.23 kJ
–133.58
101.33 kJ
Finally, the change in the internal energy during AB must be 42.9 kJ.
AB
BC
CD
DA
∆Eint
–42.9 kJ
0J
Q
0J
100 kJ
133.58 kJ
–150 kJ
W
42.9 kJ
–94.23 kJ
–133.58
101.33 kJ
Check your answer using the other internal energies.
AB
BC
CD
DA
∆Eint
–42.9 kJ
5.77 kJ
0J
–48.67 kJ
Q
0J
100 kJ
133.58 kJ
–150 kJ
W
42.9 kJ
–94.23 kJ
–133.58
101.33 kJ
The change in the internal energy
ΔE BA = −42.9 kJ
page 8
Problem 20.35
Solution
The total work done on the gas is
WBC +WDA = −(3Pi )(3Vi −Vi ) − (Pi )(Vi − 3Vi ) = −6PiVi + 2PiVi = −4PiVi
WBC +WDA = −4nRTi = −4(1 mol)(8.314
J )(273
mol⋅K
K) = −9, 078.9 J
The total change of the thermal energy is zero through a cycle, so the heat is
Q = +4PiVi = 9, 078.9 J
page 9
Problem 20.39
A glass windowpane in a home is 0.620 cm thick and has dimensions of 1 m x 2 m. On a certain
day, the temperature of the interior surface of the glass is 25 °C and the exterior surface
temperature is 0 °C.
What is the rate at which energy is transferred through the glass?
How much energy is transferred through the window in one day assuming the temperature stay
constant?
Solution
This is conduction. The two reservoirs are the interior and the exterior. The heat pipe is the
window. The rate of heat flow is
Q
A
= k ΔT = (0.8
∆t
L
3
W ) (2 m ) (25
m⋅K (0.0062 m)
°C ) = 6, 452 W
Over one day, the total heat flow is
Q
= 6, 452 W ⇒ Q = (6, 452 W )(3600 × 24 s) = 5.5742 ×108 J
∆t
page 10
Problem 20.44
At noon, the sun delivers 1,000 W of radiant power to each square meter of a blacktop road. If the
road transfers energy only by radiation, what is the stead-state temperature of the road?
Solution
The road is in thermal equilibrium. The power in is 1,000 W. Therefore, the power out must also
be 1,000 W.
P = σeAT 4 = 1000 W = (5.670 ×10 -8
W )(1)(1
m 2 ⋅K 4
m 2 )T 4
T = 364.42 K
page 11
Problem 20.61
Water in an electric tea kettle is boiling. The power absorbed by the water is 1 kW. Assuming the
pressure of vapor in the kettle equals atmospheric pressure, determine the speed of diffusion of
vapor from the kettle’s spout if this balance has a cross-sectional area of 2 cm². Model the steam
as an ideal gas.
Solution
The rate of steam mass generation is
∆m
Q
1, 000 W
=
=
= 4.4248 ×10−4
J
∆t
Lv ∆t
2.26 ×106 kg
kg
s
To convert this to a volume requires that we know how many moles this is. This will tell us what
the volume is.
∆m
1 mol steam
= (4.4248 ×10−4
∆t 18 ×10−3 kg steam
kg
)(
s
1 mol steam
18 ×10−3 kg steam
) = 2.4582 ×10−2
mol
s
The volume of this many moles of steam is
PV = nRT
⇒ V =
nRT
(2.4582 ×10−2 )(8.314)(373)
=
= 7.5235 ×10−4 m 3
P
101, 325
Every second, this volume of steam must push out of the kettle through a hole of area 2 cm2. To
get this volume, the steam must be moving at a speed of
V = 7.5235 ×10−4 m 3 = LA = vtA = v(1 s)(2 ×10−4 m 2 )
v = 3.7618
m
s
page 12
Problem 20.69
Solution
While the water is boiling, the expansion mechanism is vaporization. The rate of change of the
volume is
PV = nRT
dV
dh
d ⎛ nRT ⎞⎟
⎟
=A
= ⎜⎜
dt
dt
dt ⎜⎝ P ⎟⎟⎠
⇒
⇒
dh
1 d ⎛⎜ nRT ⎞⎟
⎟
=
⎜
dt
A dt ⎜⎝ P ⎟⎟⎠
The temperature is a constant 100 °C or 373 K. The pressure applied to the steam is one
atmosphere plus the weight of the piston and it is also constant.
P = 101, 325 Pa +
(3 kg)(9.8 m2 )
mg
s
= 101, 325 Pa +
= 102, 990 Pa
A
π(0.075 m)2
The number of moles of steam is increasing.
dh
RT dn
=
dt
PA dt
The rate at which steam is produced is
⎞⎟ dQ / dt ⎛
⎞⎟
dn dm ⎛⎜
1 mol steam
1 mol steam
⎜⎜
⎟⎟ =
⎟⎟
=
⎜⎜
⎜
dt
dt ⎝ 18 ×10−3 kg steam ⎟⎠
Lv ⎝ 18 ×10−3 kg steam ⎟⎠
dn
100 W
=
dt
2.26 ×106
J
kg
⎛
⎞⎟
1 mol steam
−3
⎜⎜
⎟
⎜⎝ 18 ×10−3 kg steam ⎟⎟⎠ = 2.4582 ×10
mol
s
This is a constant so the rate of height change is also constant.
J )(373 K)
(8.314 mol⋅K
dh
RT dn
=
=
(2.4582 ×10−3
dt
PA dt
(102, 990 Pa)π(0.075 m)2
mol )
s
= 4.1887 ×10−3 m/s
dh
= 4.19 mm/s
dt
page 13
After all of the steam boiled away, the expansion mechanism is now just heating. The rate in
which the volume increase is
PV = nRT
⇒
dV
d ⎛ nRT ⎞⎟
⎟
= ⎜⎜
dt
dt ⎜⎝ P ⎟⎟⎠
The number of moles is constant since there is only the 2 kg of steam in the volume. The number
of moles is
⎛
⎞⎟
1 mol steam
⎟⎟ = 111.11 mol
2 kg steam ⎜⎜⎜
−3
⎝ 18 ×10 kg steam ⎟⎠
The pressure is still due to the same two sources so it is also a constant.
increases. The rate of height change is
dV
nR dT
=
dt
P dt
⇒
The temperature
dh
nR dT
=
dt
PA dt
The temperature change is caused by the heat flow.
Q = mc ∆T
⇒
∆T
dQ / ∆t
=
∆t
mc
Since the heat flow rate is a constant, then the derivative is equal to the slope.
dT
∆T
Q / ∆t
100 W
=
=
=
dt
∆t
mc
(2 kg)(2010
J )
kg⋅K
= 2.4876 ×10−2
K
s
And the final result is
J )
(111.11 mol)(8.314 mol⋅K
dh
nR dT
=
=
(2.4876 ×10−2
2
dt
PA dt
(102, 990 Pa)π(0.075 m)
dh
= 12.6
dt
K)
s
= 1.2626 ×10−2
m
s
mm
s
page 14