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

26.4 :)

Step-by-step 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- (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- (313 K)/(425 K)=0.264 = 26.4 \%`