(1) Determine the heat dissipated from the refrigerator
The coefficient of performance(COP) of a refrigerator is expressed as
COP_{R} = Q_{L} /W_{net,in} = (Q_{H} 
W_{net,in})/W_{net,in}
Rearranging this equation gives
Q_{H} = (1 + COP_{R})W_{net,in} = (1 + 3.0)(1.0) =
4.0 kW
(2) Determine power generated by the heat engine
The thermal
efficiency of a heat engine is expressed as
η_{th} =
W_{net,out}/Q_{H}
where
W_{net,out} = power generated by the heat engine
Q_{H} = heat transferred to
the heat engine
Rearranging this equation yields
W_{net,out} = η_{th} Q_{H}
where Q_{H} is the heat transferred from the heatexchange
pipes to the heat engine, which is the heat dissipated from the refrigerator,
hence
Q_{H} = 4.0 kW
The thermal efficiency of a Carnot heat engine is
η_{th,rev} = 1 (273 + 20)/(273 +
30) = 3.3%
The work output from this Carnot heat engine is
W_{net,out} = η_{th,rev} Q_{H
} = (3.3%)(4.0) = 0.12 kW
The result shows the power generated from this Carnot heat engine is
only enough to light a 100 W bulb. Hence, there is little benefit to
use the heat dissipated by a refrigerator.
