Question

Heat is allowed to flow from the heat source of a heat engine at 425 K to a cold sink at 313 K. What is the efficiency of the heat engine? The answer should have three significant figures. %

Answer 1

Answer:

26.4 :)

Explanation:

Answer 2

I assume here that the engine operates following a Carnot cycle, which achieves the maximum possible efficiency. 

Under this assumption, the efficiency of the engine (so, the efficiency of the Carnot cycle) is given by
$\eta = 1- \frac{T_{cold}}{T_{hot}}$
where
$T_{cold}$ is the cold temperature
$T_{hot}$ is the hot temperature

For the engine in our problem, the cold temperature is 313 K while the hot temperature is 425 K, so the effiency of the engine is
$\eta=1- \frac{313 K}{425 K}=0.264 = 26.4 \%$