Question

The walls of a house consist of L=0.02 m thick plywood backed by insulation with the same thickness. The temperature of the inside surface of the wall (the insulation side) is T2=25 °C, while the temperature at the outside surface (the plywood side) is T1= -15 °C, both being constant. The thermal conductivities of the plywood and insulation are, respectively k1=0.09 J/(s m °C) and k2=0.03 J/(s m °C). a) Find the temperature T at the plywood-insulation interface. b) If the total surface area of the walls is 70 m^2, find the amount of heat the house loses in 24 hours. (Assuming all the heat losses are due to the thermal conductivity of the walls). c) Does the entropy of the house change? If it does, by how much? d) Does the entropy of the Universe change? If it does, by how much? e) If the cost of electric power is $0.15 per kilowatt-hour, how much does one have to pay to keep the house warm for 24 hours? 

          The walls of a house consist of L=0.02 m thick plywood backed by insulation with the same thickness. The temperature of the inside surface of the wall (the insulation side) is T2=25 °C, while the temperature at the outside surface (the plywood side) is T1= -15 °C, both being constant. The thermal conductivities of the plywood and insulation are, respectively k1=0.09 J/(s m °C) and k2=0.03 J/(s m °C).

a) Find the temperature T at the plywood-insulation interface.
b) If the total surface area of the walls is 70 m^2, find the amount of heat the house loses in 24 hours. (Assuming all the heat losses are due to the thermal conductivity of the walls).
c) Does the entropy of the house change? If it does, by how much?
d) Does the entropy of the Universe change? If it does, by how much?
e) If the cost of electric power is $0.15 per kilowatt-hour, how much does one have to pay to keep the house warm for 24 hours?
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Added by Anthony F.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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The walls of a house consist of L=0.02 m thick plywood backed by insulation with the same thickness. The temperature of the inside surface of the wall (the insulation side) is T2=25 °C, while the temperature at the outside surface (the plywood side) is T1= -15 °C, both being constant. The thermal conductivities of the plywood and insulation are, respectively k1=0.09 J/(s m °C) and k2=0.03 J/(s m °C). a) Find the temperature T at the plywood-insulation interface. b) If the total surface area of the walls is 70 m^2, find the amount of heat the house loses in 24 hours. (Assuming all the heat losses are due to the thermal conductivity of the walls). c) Does the entropy of the house change? If it does, by how much? d) Does the entropy of the Universe change? If it does, by how much? e) If the cost of electric power is $0.15 per kilowatt-hour, how much does one have to pay to keep the house warm for 24 hours?
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Transcript

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00:01 Find the temperature t at the plywood insulation interface.
00:04 So at the interface, we want to set our heat of our insulation equal to the heat for our plywood.
00:11 We have k1 times our area times delta t divided by l equal to k2 times our area times delta t over l.
00:22 We can now plug in values.
00:24 We have 0 .09 times 25 minus t equal to 0 .03, equal to 0 .03, times t minus 15.
00:33 And we find t to be 22 .5 degrees celsius.
00:40 Now for part b, we'd like to find the amount of heat the house loses in 24 hours.
00:49 Now, our q is our heat loss.
00:54 We want to plug in k times a times delta t over l.
00:59 So we plug in 0 .09 times 70 times 25 minus.
01:05 Is 22 .5 times 86 400 seconds divided by .02.
01:12 We get q to be 6 .8 times 10 to the 7 joules of heat.
01:19 Now for c we'd like to find, does the entropy of the house change? so our change in entropy is equal to negative q divided by t...
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