Question

A Carnot heat engine has an efficiency of 0.400. If it operates between a deep lake with a constant temperature of 298.0k and a hot reservoir, what is the temperature of the hot reservoir?

Answer 1

496.7 K

Step-by-step explanation:

The efficiency of a Carnot engine is given by the equation:

`\eta = 1 - (T_H)/(T_L)`

where:

`T_H` is the temperature of the hot reservoir

`T_C` is the temperature of the cold reservoir

For the engine in the problem, we know that

`\eta = 0.400` is the efficiency

`T_C = 298.0 K` is the temperature of the cold reservoir

Solving for
`T_H`, we find:

`(T_C)/(T_H)=1-\eta\\T_H = (T_C)/(1-\eta) =(298.0)/(1-0.400)=496.7 K`