Physics 9B Practice Final Problem #3

The insulation of your freezer is not perfect. When the room temperature is 300K and the temperature in the freezer is 260K, the heat flow through the insulation and into the freezer is 25 J in 1 second. a) Assuming the freezer is maximally efficient, what energy per second does the freezer compressor draw from the wall socket to maintain the temperature difference 300K and 260K? b) Your “clever” roommate says that he will run a heat engine between the inside and outside of the freezer to produce electricity to help run the freezer compressor. Suppose his engine is as efficient as possible and produces 2 watts to help run the freezer compressor. Now how much energy is needed from the wall socket each second? Explain.
Question so kindly provided by Joe Kiskis

Part A
We need to find the work needed to match the 25J/s flowing into the freezer to keep it at a constant temperature. We're going to need two equations to find this:
KCarnot = TC⁄[TH−TC]
K = QC⁄W
Solve these two equations for W and you get:
  • W = QC ⋅ [TH−TC]⁄TC
  • W = 25J ⋅ [300K - 260K]⁄260K
  • W = 3.85 J/s = 3.85 Watts
Part B
While this heat engine is producing 2 watts, it is also dumping energy into the freezer. We need to find QC for this engine.
  • eCarnot = [TH−TC]⁄TH
  • eCarnot = 40K⁄300K
  • eCarnot = 0.133
Now use this e to find QC, using |QC| = |QH| − W:
  • e = W⁄QH
  • QH = W⁄e
  • QH = 2watts⁄0.133 = 15 W
  • QC = 15W − 2W
  • QC = 13 J⁄s
Using this new QC, we can recalculated the total work needed by the Carnot refrigerator in Step A:
  • Wtotal = 38 J⁄s ⋅ 40K⁄300K = 5.85 J⁄s
    • Now remember that this is the total work, which is the sum of the works from the engine and the wall.
    • The 38J is the 25J leaking through the insulation + 13J being expelled by the heat engine.
  • Wnew = Wtotal − Wengine
  • Wnew = 5.85W − 2W
  • Wnew = 3.85W
So now realize that in the best case, adding the engine does no good. And thus, I have disproven any idea of a perpetual motion machine.

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