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.(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Carnot engines

**Physics Forums | Science Articles, Homework Help, Discussion**