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 20.7In 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?

SolutionThe amount of radiation heat loss is

Qloss = 6000 W ⋅14 hours

3600 s1 hour

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ = 3.0240×108 J

This heat comes from the temperature change of the stone.

Qstone = −3.0240×108 J = mstonecstone(Tf −Ti) = mstone(850 J

kg⋅K)(18 °C − 38 °C)

−3.0240×108 J = mstone(850 J

kg⋅K)(18 °C − 38 °C) ⇒ mstone = 1.7788×104 kg

mstone = 1.78×104 kg

page 2

Problem 20.8A 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?

SolutionThe heat flow out of the system every minute is

Qcup +Qwater = mcupcalΔT + mwcwΔT = (0.2 kg)(900 J

kg⋅K)(−1.5 K) + (0.8 kg)(4186 J

kg⋅K)(−1.5 K)

Qcup +Qwater = 5293 J

The heat flow rate is

Qcup +Qwater

Δt=

5293 J60 s

= 88.2 W

page 3

Problem 20.13Two 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?

SolutionThe number of moles of oxygen in vessel 1 is

PV = nRT ⇒ n =

PVRT

=(1.75 atm)(101,325 Pa

atm)(16.8×10−3 m3)

(8.314 Jmol ⋅K

)(300 K)= 1.1944 mol

The number of moles of oxygen in vessel 2 is

PV = nRT ⇒ n =

PVRT

=(2.25 atm)(101,325 Pa

atm)(22.4×10−3 m3)

(8.314 Jmol ⋅K

)(450 K)= 1.3650 mol

The mass of oxygen in vessel 1 is

m = 1.1944 mol

32 g1 mol

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ = 38.221 g

The mass of oxygen in vessel 2 is

m = 1.3650 mol

32 g1 mol

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ = 43.680 g

The heat flow between the oxygen in the two vessels add to zero.

Q1 +Q2 = 0

m1coxyΔT1 + m2coxyΔT2 = 0

(38.221)(Tf − 300) + (43.680)(Tf − 450) = 0

Tf = 380 K

The final pressure is

PV = nRT ⇒ P =

nRTV

=(2.5594 mol)(8.314 J

mol ⋅K)(380 K)

39.2×10−3 m3= 2.0627×105 Pa = 2.04 atm

page 4

Problem 20.17Steam 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.

SolutionHere is the summary of the results.

T (0 °C) ice (0.05 kg) T (100 °C) steam (0.01 kg)steam (0.01 kg)

melt (0 °C) 16,650 J condense (100 °C) 22,600 J22,600 J

T (0 °C) water (0.05 kg) T (100 °C) water(0.0073673 kg)

steam(0.0026327 kg)

T (100 °C) 20,930 J condense (100 °C) 5,950 J

T (28.128 °C) water (0.05 kg) T (100 °C) water (0.01 kg)water (0.01 kg)

T (40.357 °C) water (0.05 kg) T (40.357 °C) water (0.01 kg)water (0.01 kg)

For all of the ice to melt, the heat required is

Q = Lf ∆m = (3.33×105 J

kg)(0.05 kg) = 16,650 J

For all of the steam to condense, the heat required is

Q = Lv ∆m = (2.26×106 J

kg)(0.01 kg) = 22,600 J

All of the ice melt. The amount of steam that condense is

(2.26×106 J

kg)Δm = 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

kg⋅°C)(100 °C) = 20,930 J

So all of the steam go to water. The cold water warms to

(0.05 kg)(4186 J

kg⋅°C)(Tf − 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)ice (0.05 kg) T (100 °C) steam (0.001 kg)

melt (0 °C) 16,650 J16,650 J condense (100 °C) 2,260 J

T (0 °C) water(0.0067868 kg)

ice(0.043213 kg) T (100 °C) water (0.001 kg)

melt (0 °C) 14,390 J T (0 °C) 418.6 J

T (0 °C) water(0.0080439 kg)

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

kg)Δm = 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

kg⋅°C)(100 °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

kg)Δm = 418.6 J ⇒ ∆m = 0.0012571 kg

The total amount of ice that melts is 0.0080439 g.

page 6

Problem 20.30

SolutionPut the numbers in a table again starting with the givens.

∆Eint Q WABBC 100 kJCDDA –150 kJ

The change in the internal energy is zero during CD. The heat is zero during AB.

∆Eint Q WAB 0 JBC 100 kJCD 0 JDA –150 kJ

The work done during BC is

W = −(3)(101,325 Pa)(0.31 m3) = −94.232 kJ

The work done during DA is

W = −(1)(101,325 Pa)(−1 m3) = 101.325 kJ

The work done during CD is

W = −nRTCD ⋅ ln

VD

VC

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

= −PCVC ⋅ lnVD

VC

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

= −(3)(101,325)(0.4) ⋅ ln1.20.4

⎛

⎝⎜⎜⎜⎞

⎠⎟⎟⎟⎟ = −133.58 kJ

page 7

∆Eint Q WAB 0 JBC 100 kJ –94.23 kJCD 0 J –133.58DA –150 kJ 101.33 kJ

This means the heat during CD is +133.58 kJ making total heat through the entire cycle 83.58 kJ.

∆Eint Q WAB 0 JBC 100 kJ –94.23 kJCD 0 J 133.58 kJ –133.58DA –150 kJ 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 Q WAB 0 J 42.9 kJBC 100 kJ –94.23 kJCD 0 J 133.58 kJ –133.58DA –150 kJ 101.33 kJ

Finally, the change in the internal energy during AB must be 42.9 kJ.

∆Eint Q WAB –42.9 kJ 0 J 42.9 kJBC 100 kJ –94.23 kJCD 0 J 133.58 kJ –133.58DA –150 kJ 101.33 kJ

Check your answer using the other internal energies.

∆Eint Q WAB –42.9 kJ 0 J 42.9 kJBC 5.77 kJ 100 kJ –94.23 kJCD 0 J 133.58 kJ –133.58DA –48.67 kJ –150 kJ 101.33 kJ

The change in the internal energy

ΔEBA = −42.9 kJ

page 8

Problem 20.35

SolutionThe 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

mol ⋅K)(273 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 20.39A 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?

SolutionThis is conduction. The two reservoirs are the interior and the exterior. The heat pipe is the window. The rate of heat flow is

Q∆t

= kALΔT = (0.8 W

m⋅K)

(2 m3)(0.0062 m)

(25 °C) = 6,452 W

Over one day, the total heat flow is

Q∆t

= 6,452 W ⇒ Q = (6, 452 W )(3600×24 s) = 5.5742×108 J

page 10

Problem 20.44At 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?

SolutionThe 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

m2⋅K 4 )(1)(1 m2)T 4

T = 364.42 K

page 11

Problem 20.61Water 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.

SolutionThe rate of steam mass generation is

∆m∆t

=Q

Lv ∆t=

1,000 W

2.26×106 Jkg

= 4.4248×10−4 kgs

To convert this to a volume requires that we know how many moles this is. This will tell us what the volume is.

∆m∆t

1 mol steam

18×10−3 kg steam= (4.4248×10−4 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 =

nRTP

=(2.4582×10−2)(8.314)(373)

101,325= 7.5235×10−4 m3

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 m3 = LA = vtA = v(1 s)(2×10−4 m2)

v = 3.7618 m

s

page 12

Problem 20.69

SolutionWhile the water is boiling, the expansion mechanism is vaporization. The rate of change of the volume is

PV = nRT ⇒

dVdt

= Adhdt

=ddt

nRTP

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ ⇒

dhdt

=1A

ddt

nRTP

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

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 +

mgA

= 101,325 Pa +(3 kg)(9.8 m

s2)

π(0.075 m)2= 102,990 Pa

The number of moles of steam is increasing.

dhdt

=RTPA

dndt

The rate at which steam is produced is

dndt

=dmdt

1 mol steam

18×10−3 kg steam

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

=dQ / dt

Lv

1 mol steam

18×10−3 kg steam

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

dndt

=100 W

2.26×106 Jkg

1 mol steam

18×10−3 kg steam

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

= 2.4582×10−3 mols

This is a constant so the rate of height change is also constant.

dhdt

=RTPA

dndt

=(8.314 J

mol ⋅K)(373 K)

(102,990 Pa)π(0.075 m)2(2.4582×10−3 mol

s) = 4.1887×10−3 m/s

dhdt

= 4.19 mm/s

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 ⇒

dVdt

=ddt

nRTP

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

The number of moles is constant since there is only the 2 kg of steam in the volume. The number of moles is

2 kg steam

1 mol steam

18×10−3 kg steam

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

= 111.11 mol

The pressure is still due to the same two sources so it is also a constant. The temperature increases. The rate of height change is

dVdt

=nRP

dTdt

⇒ dhdt

=nRPA

dTdt

The temperature change is caused by the heat flow.

Q = mc∆T ⇒

∆T∆t

=dQ /∆t

mc

Since the heat flow rate is a constant, then the derivative is equal to the slope.

dTdt

=∆T∆t

=Q /∆t

mc=

100 W(2 kg)(2010 J

kg⋅K)

= 2.4876×10−2 Ks

And the final result is

dhdt

=nRPA

dTdt

=(111.11 mol)(8.314 J

mol ⋅K)

(102,990 Pa)π(0.075 m)2(2.4876×10−2 K

s) = 1.2626×10−2 m

s

dhdt

= 12.6 mms

page 14