Discussion

An ideal Carnot engine has an efficiency of 40%. The ratio of the temperature of the sink to that of the source is

Answer: B

The ratio is 0.6.

For a Carnot engine:

\[ \begin{aligned} \eta &= 1 - \frac{T_{2}}{T_{1}} = 0.4 \\ \frac{T_{2}}{T_{1}} &= 1 - 0.4 = 0.6 \end{aligned} \]

0.4 is the efficiency itself, not the temperature ratio.

So if the source is at 500 K, the sink is at 300 K. The ratio is always \(1 - \eta\), whichever temperatures are chosen.

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