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

Answer:

496.7 K

Explanation:

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

$\eta = 1 - \frac{T_H}{T_L}$

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:

$\frac{T_C}{T_H}=1-\eta\\T_H = \frac{T_C}{1-\eta} =\frac{298.0}{1-0.400}=496.7 K$