In a cyclic heat engine operating between a source temperature of 600°C & a sink temperature of 20°C, the least rate of heat rejection per kW net output of the engine is

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The least rate of heat rejection per kW net output of the engine is determined by the maximum efficiency, which is given by the Carnot efficiency formula: $\eta = 1 - \frac{T_{sink}}{T_{source}}$, where temperatures are in Kelvin. Converting the given temperatures to Kelvin: $T_{source} = 600 + 273 = 873K$ and $T_{sink} = 20 + 273 = 293K$. Thus, $\eta = 1 - \frac{293}{873} = 0.664$. The efficiency tells us that 66.4% of the heat input is converted to work, and the rest (33.6%) is rejected as heat. For a net output of 1 kW (which represents 66.4% of the input), the heat rejected is $1 kW / 0.664 = 1.506 kW$ of input, and the rejection is $1.506 kW - 1 kW = 0.506 kW$.

In a Cyclic Heat Engine Operating Between a Source Temperature