A Carnot engine operates between a source temperature of 527°C and a sink temperature of 27°C. What is the efficiency of the engine?

The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$, where temperatures must be in Kelvin. $T_1 = 527 + 273 = 800$ K and $T_2 = 27 + 273 = 300$ K. Therefore $\eta = 1 - \frac{300}{800} = 1 - 0.375 = 0.625 = 62.5\%$. Option A results from using Celsius temperatures directly (500/2000), option B from incorrectly halving the temperature ratio, and option D from computing $(T_1 - T_2)/T_1$ using Celsius values.