Carnot Heat Pump: Solving |Q_c|/|Q_h|=T_c/T_h

[frankwilson]

A Carnot engine uses hot and cold reservoirs that have temperatures of 1684 and 842 K, respectively. The input heat for this engine is |QH|. The work delivered by the engine is used to operate a Carnot heat pump. The pump removes heat from the 842-K reservoir and puts it into a hot reservoir at a temperature T`. The amount of heat removed from the 842-K reservoir is also |QH|. Find the temperature T`.


|Q_c|/|Q_h|=T_c/T_h
|Q_h|=|W| + |Q_c|


I'm having trouble visualizing this problem. Are there two separate engines? Doing a little rearranging, I was able to get down to |W| = 1/2 |Q_h|. I figure that since there are no values for either heat value or work that they cancel out. I'm just not sure how to proceed. I worked it one way and got my final T` to be 1684 K, but I don't feel too confident about it. Anyone out there know where I should start or if I'm even on the right track?

 

[Andrew Mason]

A Carnot engine uses hot and cold reservoirs that have temperatures of 1684 and 842 K, respectively. The input heat for this engine is |QH|. The work delivered by the engine is used to operate a Carnot heat pump. The pump removes heat from the 842-K reservoir and puts it into a hot reservoir at a temperature T`. The amount of heat removed from the 842-K reservoir is also |QH|. Find the temperature T`.|Q_c|/|Q_h|=T_c/T_h
|Q_h|=|W| + |Q_c|I'm having trouble visualizing this problem. Are there two separate engines?

There are two Carnot devices. One is a heat engine, the other is a heat pump.

Doing a little rearranging, I was able to get down to |W| = 1/2 |Q_h|. I figure that since there are no values for either heat value or work that they cancel out. I'm just not sure how to proceed. I worked it one way and got my final T` to be 1684 K, but I don't feel too confident about it. Anyone out there know where I should start or if I'm even on the right track?

Write out the equation for the COP of the heat pump as a function of Tc and T': COP = W/Qc

Since W = Qh/2 and Qc = Qh, that leaves you with an equation with only one unknown: T'.

AM

 

[Vespa71]

Or for simplicity maybe just Tc/Th=Qc/Qh, then rearrange and substitute. By the way, AM, COP-figures for consumer-info usually are larger than 1, calculated Qc/W for coolers, aren't they? Anyway, no need to bring in Cop or efficiency as long as we know enough T's and Q's.

 

[Andrew Mason]

Or for simplicity maybe just Tc/Th=Qc/Qh, then rearrange and substitute. By the way, AM, COP-figures for consumer-info usually are larger than 1, calculated Qc/W for coolers, aren't they?

Yes - a slip there. COP = output/input = heat removed/work input = Qc/W. Thanks.

AM