For a certain ideal Carnot engine,the hot reservoir is 35°C higher than the cold reservoir.If this engine is to have an efficiency of 20%,what must be the temperature of the hot reservoir?
A)70.0 K
B)140 K
C)175 K
D)210 K
E)245 K

Question

Quizplus · Accepted Answer

The efficiency of a Carnot engine is given by the equation:$$e=\frac{T_H-T_C}{T_H}$$where $T_H$ is the temperature of the hot reservoir and $T_C$ is the temperature of the cold reservoir.We are given that the efficiency of the engine is 20%, or 0.2. We are also given that the hot reservoir is 35°C higher than the cold reservoir. This means that:$$T_H=T_C+35$$Substituting this into the equation for efficiency, we get:$$0.2=\frac{T_C+35-T_C}{T_C+35}$$Simplifying, we get:$$0.2=\frac{35}{T_C+35}$$Solving for $T_C$, we get:$$T_C=175$$Therefore, the temperature of the hot reservoir is:$$T_H=T_C+35=175+35=210$$So, the correct answer is C) 175 K.

For a Certain Ideal Carnot Engine,the Hot Reservoir Is 35°C