A Carnot heat engine is to be designed with an efficiency of 81%.If the low temperature reservoir is 25 °C,what is the temperature of the "hot" reservoir?

Question

Quizplus · Accepted Answer

The efficiency of a Carnot engine is given by:

η = 1 - T_low / T_high

where T_low is the temperature of the low temperature reservoir, and T_high is the temperature of the high temperature reservoir.

We know that the efficiency is 81%, so we can write:

0.81 = 1 - 298 / T_high

Solving for T_high, we get:

T_high = 1 / (1 - 0.81) * 298 = 1574 K = 1300 °C

Therefore, the temperature of the hot reservoir is 1300 °C. The answer is (D).

A Carnot Heat Engine Is to Be Designed with an Efficiency