Entropy in the Real World: Refrigerators — Question 7

Section 20-5 · Physics pages 3324–3325

Question 7 An open refrigerator warms the insulated room that contains it An operating refrigerator stands with its door open in an insulated room. It receives 200 W of electrical power from outside and has instantaneous cooling rate 600 W. Both its cold and warm heat exchangers lie inside the room. The combined room and apparatus heat capacity is 10000 J/K, with negligible other energy storage. open refrigerator 200 W room boundary Schematic; labels, not drawing lengths, specify the data. (a) Find heat rejection rate and net room heating rate. (b) Find the initial temperature-rise rate and the time for a 1.00 K rise if power and heat capacity remain constant. Would increasing cooling coefficient at the same electrical power change that net heating?
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Question 7 - Solution An open refrigerator warms the insulated room that contains it open refrigerator 200 W room boundary (a) Method and calculation ˙Qh = ˙Qc + ˙W = 600 + 200 = 800 W. ˙Qnet,room = ˙Qh −˙Qc = 200 W. (b) Interpretation and further calculation C dT dt = 200 ⇒dT dt = .0200 K/s, t = C(1.00) 200 = 50.0 s. ˙Qh −˙Qc = ˙W for any cyclic coefficient. Internal heat relocation cancels at the room boundary; the external electrical input remains. Physical check The conclusion depends on both heat exchangers being inside the same insulated boundary.

Question 7

An open refrigerator warms the insulated room that contains it

An operating refrigerator stands with its door open in an insulated room. It receives 200 W of electrical power from outside and has instantaneous cooling rate 600 W. Both its cold and warm heat exchangers lie inside the room. The combined room and apparatus heat capacity is 10000 J/K, with negligible other energy storage.

Diagram shown in the original SVG worksheet.

  1. Find heat rejection rate and net room heating rate.

  2. Find the initial temperature-rise rate and the time for a 1.00 K rise if power and heat capacity remain constant. Would increasing cooling coefficient at the same electrical power change that net heating?

Question 7 - Solution

An open refrigerator warms the insulated room that contains it

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Q̇h=Q̇c+Ẇ=600+200=800W.\dot Q_h=\dot Q_c+\dot W=600+200=800\,\mathrm W.Q̇net,room=Q̇h−Q̇c=200W.\dot Q_{\mathrm{net,room}}=\dot Q_h-\dot Q_c=200\,\mathrm W.

(b) Interpretation and further calculation

CdTdt=200⇒dTdt=.0200K/s,t=C(1.00)200=50.0s.C\frac{dT}{dt}=200\ \Rightarrow\frac{dT}{dt}=.0200\,\mathrm{K/s},\quad t=\frac{C(1.00)}{200}=50.0\,\mathrm s.Q̇h−Q̇c=Ẇfor any cyclic coefficient.\dot Q_h-\dot Q_c=\dot W\quad\text{for any cyclic coefficient}.Internal heat relocation cancels at the room boundary; the external electrical input remains.

Physical check

The conclusion depends on both heat exchangers being inside the same insulated boundary.