| ToC, Chapter 17 |Course Calendar|

**D17.1** A goose-down sleeping bag has a surface area of 2.25 m^{2}
and is filled with a layer of down 5 cm thick. What is the heat-transfer rate
(H_{cd}) through it from a person with skin temperature of 35°C
to the outside air at -5°C. How does this rate compare with the body’s
minimal metabolism rate, about 100 W?

The thermal conductivity of goose down isK = 0.023 W/m C° We can now apply the conduction heat-transfer equation, Eq 13.1,

H _{cd}= [K A (T_{2}- T_{1})] / LFrom the problem, we know the other pieces of this equation,

DT = T _{2}- T_{1}= 35°C - (- 5°C) = 40 C°

A = 2.25 m^{2}

L = 5 cm = 0.05 m

H_{cd}= [K A (T_{2}- T_{1})] / L

H_{cd}= [(0.023 W/m C°)(2.25 m^{2})(40 C°)]/0.05 m =

H_{cd}= 41 WThis is about half of the body’s normal metabolism so this sleeping bag should keep our camper a “happy camper” (or, at least, a warm camper).

(Hmmm, . . . go ahead and use e = 1.00, even in this case).

To find the temperature, we can use the Radiation equation, Eq 13.4,R _{emit}= e A T^{4}We will use e 1.0, even for this. s is the Stefan-Boltzman constant,

= 5.67 x 10 ^{- 8}W/_{m}2_{K}4The surface area of a sphere is

A = (4) r ^{2}For the

innersurface,A _{inner}= (4) (0.200 m)^{2}= 0.50265 m^{2}For the

outersurface,A _{out}= (4) (0.205 m)^{2}= 0.5281 m^{2}The temperatures T must be

absolutetemperatures, measured on the Kelvin scale.

Now we are ready to solve for the temperatursT ^{4}= R_{emit}/ e A_{ }T_{inner}^{4}=^{ }100 W/[(1.0)(5.67 x 10^{ - 8}W/m^{2}K^{4})(0.50265 m^{2})] = 3.509 x 10^{ 9}K^{4}

T_{inner}= 243 K_{ }T_{outer}^{4}= 100 W/[(1.0)(5.67 x 10^{ - 8}W/m^{2}K^{4})(0.5281 m^{2})] = 3.3396 x 10^{ 9}K^{4}

T_{outer}= 240 K

Our use of e = 1.0 , for the emissivity, makes our actual numbers rather “iffy”. But thepointof this problem was to show that theinnersurface will be at a higher temperature than theoutersurface. That is certainly true for our calculation. Changing the value of the emissivity e may change our actual numbers but we will still find T_{inner}> T_{outer}.

N and S mark the poles, the axis of rotation. Z is the zenith, the point directly overhead, 48.5° from the equator (48.5° north latitude).

Summer:

is the angle from directly overhead (the zenith) to the Sun. Therefore, the angle between the horizon and the Sun will be 90° - = 90° - 25° = 65°. The Sun will be 65° above the horizon at the summer solstice.

Winter:

is again the angle between directly overhead (the zenith) and the Sun. Therefore, the Sun is 90° - 72° = 18° above the horizon at the winter solstice.

The first part of this questions requires exactly the same reasoning, using exactly the same kind of diagrams, as question 13.21.N and S mark the poles, the axis of rot ation. Z is the zenith, the point directly overhead, 45° from the equator (45° north latitude).

Summer:If the Sun is 21.5° from the zenith (the position directly overhead) then it must be 90° - 21.5° = 68.5° above the horizon.

Winter:If the Sun is 68.5° from the zenith, then it must be 90° - 68.5° = 21.5° above the horizon.

Now we can look at the shadow cast by the eave on these two days:

Summer:The length of the shadow, S in the diagram, is the

oppositeside of a 68.5° right triangle (85 cm is the adjacent side).tan = ^{opp}/_{adj}

tan 68.5° =^{S}/_{ 85 cm}_{ }S = (85 cm)(tan 68.5°)

S = (85 cm)(2.54) = 216 cm = 2.16 cm

Winter:The length of the shadow, S in the diagram, is the

oppositeside of a 21.5° right triangle (85 cm is the adjacent side).tan = ^{opp}/_{adj}

tan 21.5° =^{S}/_{ 85 cm}

S = (85 cm)(tan 21.5°)

S = (85 cm)(0.393) = 33.5 cm

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CHANGE NEEDED

********************************************************D17.5Calculate the R value of an insulated wall like that in Figure 13.30 if 2 x 6’s are used for wall studs instead of 2 x 4’s. The larger board allow 5.5 in of fiberglass insulation instead of 3.5 in.By what percentage does this increase in thickness change the R value?

How does the increased thickness change the heat loss calculated in Example 13.14?

The R-values for the components of the wall in Figure 13.30 are given in Table 13.4, on page 481. By increasing the fiberglass insulation from 3.5 inches to 5.5 inches, the R-value for this insulation increases from 10.90 to 18.80, according to Table 13.3, on page 480. This increases the total R-value from 14.33 to22.23.

This is an increase of^{7.9}/_{14.33}= 0.55 =55%for the R-value of the wall.

In terms of R-values, the conduction heat transfer rate is given by equation 13.9,H _{cd}= A (T_{2}- T_{1}) / R = (^{1}/_{R}) [A (T_{2}- T_{1})]In Example 13.14, we found the conduction heat transfer rate to be

H _{cd}(un) = 3.8 kWfor the

uninsulated wall,H _{cd}(3.5) = 1.2 kWfor the insulated wall with 2x4’s and 3.5 in of insulation

Now, with 2x6’s and 5.5 in of insulation,H _{cd}(5.5) = (^{1}/_{22.23})[(960)(60)] = 2591 Btu/h = 0.76 kWThis is a reduction of heat flow by

H = 1.18 kW - 0.76 kW = 0.42 kW

H / H =^{0.42 kW}/_{1.18 kW}= 0.36 =36%

**
**

This requires an application of Newton’s Law of Cooling,T(t) = T _{sur}+ T e^{- t/}T is the

initialtemperature difference of the turkey and its surroundings;T = 85°C - 25°C = 60 C° Knowing that it cools from 85°C to 80°C allows us to solve for the “time constant” t in this equation;

T(t) = T _{sur}+ T e^{-t/}

T(10 min) = 80°C = 25°C + (60 C°) e^{-( 10 min)/}

55°C = (60 C°) e^{-}^{ (10 min}^{)/}^{}

^{55°C}/_{60°C}= e^{- (10 min)/}

^{55}/_{60}= e^{- (10 min)/t}^{}

0.9167 = e^{- (10 min)/}^{}

e^{- (10 min)/}^{}= 0.9167

ln[e^{-}^{ (10 min)/}^{}] = ln[0.9167]

- (10 min)/^{}= -0.0870

10 min/^{}= 0.0870

^{}= (10 min)/0.0870

^{}= 115 minNow we know the time constant t and we can use Newton’s Law of Cooling to go back and solve for t, the time, when T, the temperature, is 55°C.

T(t) = T

_{sur}+ T e^{- t/}^{}

T(t) = 55°C = 25°C + (60 C°) e^{-t /(115 min)}

55°C = 25°C + (60 C°) e^{-}^{ t/(115 min)}

30°C = (60 C°) e^{- t/(115 min)}

^{30°C}/_{60°C}= e^{- t/(115 min)}

^{30}/_{60}= e^{-}^{ t /(115 min)}

0.50 = e^{- t/(115 min)}

ln[0.50] = ln[e^{- t/(115 min)}]

-0.693 = -t/(115 min)

0.693 = t/(115 min)

t = (0.693)(115 min)

t = 78 min

D17.7If concrete roadway sections are poured butted against each other when the temperature is 15°C, what will the thermal stress be when the temperature reaches 40°C? Young’s modulus for concrete is about 20 x 10^{9}Pa.

The thermal stress is given by^{F}/_{A}= Y TFrom the text, we have the coefficient of linear expansion for concrete,

= 12 x 10 ^{-6}(C°)^{- 1}

^{F}/_{A}= Y T

^{F}/_{A}= [12 x 10^{-6}(C° )^{ - 1 }] [20 x 10^{9}Pa] [25 C°]

^{F}/_{A}= 6 x 10^{6}Pa

D17.8If 0.60 kg of boiling hot coffee is poured into a 0.250-kg steel camping cup initially at 20°C, what is the final temperature of the cup-coffee system?Q _{1}+ Q_{2}= 0

Q_{1}= Q_{cup}= c_{cup}m_{cup}T_{cup}= c_{steel }m_{steel}T_{steel}From Table 12.2, p 438, we find that the specific heat of steel is c

_{steel}= 450^{J}/_{kg C°}Q _{1}= c_{steel }m_{steel}T_{steel}= ( 450^{J}/_{kg C°}) (0.250 kg) (T_{f}- 20°C)

Q_{2}= Q_{coffee}= c_{coffee}m_{coffee}T_{coffee}= c_{water }m_{water}T_{water}

Q_{2}= c_{water }m_{water}T_{water}= (4186^{J}/_{kg C°}) (0.60 kg) (T_{f}- 100°C)where we have taken the specific heat of coffee to be the same as the specific heat of water,

c

_{coffee}= c_{water}= 4186^{J }/_{kg C°}.Now we can put these all together as

Q _{1}+ Q_{2}= 0

( 450^{J}/_{kg C°}) (0.250 kg) (T_{f}- 20°C) + (4186^{J}/_{kg C°}) (0.60 kg) (T_{f}- 100°C) = 0

112.5 (^{J}/_{C°}) T_{f}- 2,250 J + 2,512(^{J}/_{C°}) T_{f}- 251,200 J = 0

112.5 (^{J}/_{C°}) T_{f}+ 2,512(^{J}/_{C°}) T_{f}= 2,250 J + 251,200 J

[ 112.5 + 2,512 ](^{J}/_{C°}) T_{f}= [ 2,250 + 251,200 ] J

2,624.5 T_{f}= 253,450 °C

T_{f}=^{253,450 °C}/_{2,624.5}

T_{f}= 96.6°C

D17.9A 0.050-kg ice cube, initially at - 5.0°C, is placed in 0.30 kg of water at 25°C. What is the final temperature of the water? Or, if melting is not complete, how much ice remains unmelted at thermal equilibrium? Assume no heat is lost to the environment.

First, assume all the ice melts, so T_{f}> 0°CQ _{1}+ Q_{2}= 0

Q_{1}= heat lost by the water

Q_{1}= c_{water}m_{water}T_{water}= (4186^{J}/_{kg C°})(0.30 kg)(T_{f}- 25°C)

Q_{1}= (4186^{J}/_{kg C°})(0.30 kg)(T_{f}- 25°C) = 1,256 (^{J}/_{C°}) T_{f}- 31,395 JQ

_{2}= heat gained by ice

Q_{2}= heat gained in raising ice to 0°C +

+ heat gained in melting ice +

+ heat gained in raising the melted ice (now water) to temperature T_{f}

Q_{2}= c_{ice}m_{ice}T_{1}+ L_{f,ice}m_{ice}+ c_{water}m_{ice}T_{2}

Q_{2}= [2090^{J}/_{kg C°}][0.05 kg][0°C - (- 5°C)] +

+ [3.33 x 10^{5}^{J}/_{kg}][0.05 kg] +

+ [4186^{J}/_{kg C°}][0.05 kg][T_{f}- 0°C]

Be very careful with these temperature changes!Q _{2}= 522.5 J + 16,650 J + 209.3 (^{J}/_{C°}) T_{f}- 0

Q_{2}= 17,172.5 J + 209.3 (^{J}/_{C°}) T_{f}Now we are ready to go back to

Q _{1}+ Q_{2}= 0

1,256 (^{J}/_{C°}) T_{f}- 31,395 J + 17,172.5 J + 209.3 (^{J}/_{C°}) T_{f}= 0

1,256 (^{J}/_{C°}) T_{f}+ 209.3 (^{J}/_{C°}) T_{f }= 31,395 J - 17,172.5 J

1,465 (^{J}/_{C°}) T_{f}= 14,222 J

T_{f}= [^{14,222}/_{1,465}]°C

T_{f}= 9.7°CAnd this is a reasonable answer.

D17.10A 0.15-kg ice cube initially at -15°C is placed in 0.30 kg of water at 25°C. What is the final temperature of the water? Or, if melting is not complete, how much ice remains unmelted at thermal equilibrium? Assume no heat is lost to the environment.

This should look very similar to the previous question or problem. Now we have a larger and colder ice cube.

First, assume all the ice melts, so T_{f}> 0°CQ _{1}+ Q_{2}= 0

Q_{1}= heat lost by the water

Q_{1}= c_{water}m_{water}T_{water}= (4186^{J}/_{kg C°})(0.30 kg)(T_{f}- 25°C)

Q_{1}= (4186^{J}/_{kg C°})(0.30 kg)(T_{f}- 25°C) = 1,256 (^{J}/_{C°}) T_{f}- 31,395 J

Q_{2}= heat gained by ice

Q_{2}= heat gained in raising ice to 0°C +

+ heat gained in melting ice +

+ heat gained in raising the melted ice (now water) to temperature T_{f}

Q_{2}= c_{ice}m_{ice}T_{1}+ L_{f,ice}m_{ice}+ c_{water}m_{ice}T_{2}

Q_{2}= [2090^{J}/_{kg C°}][0.15 kg][0°C - (- 15°C)] +

+ [3.33 x 10^{5}^{J}/_{kg}][0.15 kg] +

+ [4186^{J}/_{kg C°}][0.15 kg][T_{f}- 0°C]

Be very careful with these temperature changes!Q _{2}= 4,702 J + 49,950 J + 628(^{J}/_{C°})T_{f}- 0

Q_{2}= 54,652 J + 628(^{J}/_{C°})T_{f}

Q_{1}+ Q_{2}= 0

1,256 (^{J}/_{C°}) T_{f}- 31,395 J + 54,652 J + 628 (^{J}/_{C°})T_{f}= 0

[1,256 + 628](^{J}/_{C°})T_{f}+ [- 31,395 + 54,652]J = 0

1,884](^{J}/_{C°})T_{f}+ 23,257J = 0

1,884](^{J}/_{C°})T_{f}= - 23,257J

T_{f}= [^{- 23,257}/_{1,884}]°C

T_{f}= - 12.3°CAnd this is

with our assumption that Tinconsistent_{f}> 0. Therefore, we have to look at another possibility. This time, we shall assume T_{f}= 0°C and only m_{melt}of the ice melts. Our calculation for Q_{1}is almost the same as before but our calculation for Q_{2}is quite different.Q _{1}= heat lost by the water

Q_{1}= c_{water}m_{water}T_{water}= (4186^{J}/_{kg C°})(0.30 kg)(0°C - 25°C)

Q_{1}= (4186^{J}/_{kg C°})(0.30 kg)(- 25°C) = - 31,395 J

Q_{2}= heat gained by ice

Q_{2}= heat gained in raising ice to 0°C + heat gained in melting some ice

Q_{2}= c_{ice}m_{ice}T_{1}+ L_{f,ice}m_{melt}

Q_{2}= [2090^{J}/_{kg C°}][0.15 kg][0°C - (- 15°C)] + [3.33 x 10^{5}^{J}/_{kg}][m_{melt}]

Q_{2}= 4,702 J + 3.33 x 10^{5}(^{J}/_{kg})m_{melt}

Q_{1}+ Q_{2}= 0

- 31,395 J + 4,702 J + 3.33 x 10^{5}(^{J}/_{kg})m_{melt}= 0

- 26,693 J + 3.33 x 10^{5}(^{J}/_{kg})m_{melt}= 0

3.33 x 10^{5}(^{J}/_{kg})m_{melt}= 26,693 J

m_{melt}= [^{26,693}/_{333,000}] kg

m_{melt}= 0.080 kgSince we started with 0.150 kg, this is a reasonable answer. There will remain 0.070 kg of ice that has not melted.

D17.11The inside diameter of a steel lid and the outside diameter of a glass peanut butter jar are both exactly 11.50 cm at room temperature, 21.0°C. If the lid is stuck and you run 80°C hot water over it until the lid and the jar top both come to 80°C, what will the new diameters be?

For the steel lid, thechangein the diameter, d, isd = d _{o}T = [12 x 10^{- 6}(C°)^{-1}] [11.50 cm] [59 C°] = 0.0081 cmso the new diameter is

d _{new}= d_{o}+ d = 11.5081 cmFor the glass jar, the

changein the diameter, Dd, isd = d _{o}T = [10 x 10^{- 6}(C°)^{-1}] [11.50 cm] [59 C°] = 0.0068 cmso the new diameter is

d _{new}= d_{o}+ d = 11.5068 cmNow the

differencein the two diameters is 0.0013 cm. Thatmaybe enough to break the lid loose and let us make our peanut butter sandwich!

D17.12If concrete roadway sections are poured butted against each other when the temperature is 15°C, what will the thermal stress be when the temperature reaches 40°C? Young’s modulus for concrete is about 20 x 10^{9}Pa.

The thermal stress is given by^{F}/_{A}= Y TFrom the text, we have the coefficient of linear expansion for concrete,

= 12 x 10 ^{-6}(C°)^{- 1}

^{F}/_{A}= Y T

^{F}/_{A}= [12 x 10^{-6}(C° )^{ - 1 }] [20 x 10^{9}Pa] [25 C°]

^{F}/_{A}= 6 x 10^{6}Pa

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(c) 2005, Doug Davis; all rights reserved