(1) Determine the maximum thermal efficiency
Assuming the power plant is a Carnot heat engine, which works between
a source at temperature 85 ^{o}C, and a sink at temperature 35 ^{o}C.
η_{th,rev} =
1  (35+273)/(85+273) = 14%
(2) Determine the maximum power generated by the pond
The definition of the thermal efficiency of a heat engine is
η_{th} =
W_{net,out}/Q_{in}
Rearranging this equation gives
W_{net,out} = Q_{in }η_{th}
If the efficiency of this heat engine equals the Carnot efficiency,
and the heat absorbed by the salt water is totally input to the heat
engine, the output net work can achieve its maximum value.
W_{net,max} = Q_{H}η_{th,rev}
The bottom area of the pond is
A_{pond} = (10) (10) = 100
m^{2}
Heat absorbed by the salt water in the pond bottom area is
Q_{H} = (0.3) (100) = 30
kW
Hence, the maximum net work output by the heat engine equals
W_{net,out,max} = Q_{in }η_{th,rev} =
( 30)(0.14) = 4.2 kW
