Answer
The maximum possible efficiency of a heat engine is given by the Carnot efficiency, which is:
η_carnot = 1 - T_cold / T_hot
where T_cold is the temperature of the cold reservoir and T_hot is the temperature of the hot reservoir.
In the winter, when the river temperature is 0°C, the maximum possible efficiency of the power plant is:
η_carnot = 1 - (273.15 + 0) K / 570 K = 0.52 or 52%
In the summer, when the river temperature is 25°C, the maximum possible efficiency of the power plant is:
η_carnot = 1 - (273.15 + 25) K / 570 K = 0.43 or 43%
These efficiencies are the theoretical maximum efficiencies that the power plant could achieve. In reality, the actual efficiency will be lower due to various sources of energy loss, such as friction in the turbine, heat loss through the pipes, and imperfect heat transfer between the reactor and the steam.
Work Step by Step
The maximum possible efficiency of a heat engine is given by the Carnot efficiency, which is:
η_carnot = 1 - T_cold / T_hot
where T_cold is the temperature of the cold reservoir and T_hot is the temperature of the hot reservoir.
In the winter, when the river temperature is 0°C, the maximum possible efficiency of the power plant is:
η_carnot = 1 - (273.15 + 0) K / 570 K = 0.52 or 52%
In the summer, when the river temperature is 25°C, the maximum possible efficiency of the power plant is:
η_carnot = 1 - (273.15 + 25) K / 570 K = 0.43 or 43%
These efficiencies are the theoretical maximum efficiencies that the power plant could achieve. In reality, the actual efficiency will be lower due to various sources of energy loss, such as friction in the turbine, heat loss through the pipes, and imperfect heat transfer between the reactor and the steam.
