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Carnot engines

  • Thread starter physicsss
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  • #1
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The steam engine which operates between 500°C and 200°C is run in reverse. How long would it take to freeze a tray of a dozen 37 g compartments of liquid water at room temperature (20°C) into a dozen ice cubes at the freezing point, assuming that it takes 300 W of input electric power to run it? Assume ideal (Carnot) behavior.

The Q that is required to freeze the cube is (37/1000)kg*(4186J/K)(20K)+(37/1000)kg(3.33 x 10^5)

The coefficient of performance of ice is equal to TL/(TH - TL)=QL/(QH-QL), so pluging it the stuff, I get (200+273)/((500+273)-(200+273))= QL/ (300-QL), so I solved for QL and divide the Q needed to melt ice by QL. But my answer is wrong. Why? The 300W is QH, right? :cry:
 

Answers and Replies

  • #2
Astronuc
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The Q that is required to freeze the cube is (37/1000)kg*(4186J/K)(20K)+(37/1000)kg(3.33 x 10^5)
First of all, it seems you forgot to multiply by 12, since the problem considers a dozen 37 gram compartments.

Then to freeze one has to reduce the temperature by 20°C and then remove the heat of fusion. It appears you have considered this.
 
  • #3
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physicsss said:
assuming that it takes 300 W of input electric power to run it? Assume ideal (Carnot) behavior.
Do you think this is W(work) or QH?
 
  • #4
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And thank you for pointing that out. :)
 

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