A Carnot refrigerator operates between energy reservoirs at 0℃ and 250℃. A 2.4-cm-diameter, 50-cm-long copper bar connects the two energy reservoirs. At what rate, in W, must work be done on the refrigerator to remove heat from the cold reservoir at the same rate that it arrives through the copper bar?

Key Concepts

Carnot Refrigerator

A Carnot refrigerator is an idealized refrigeration cycle that operates between two thermal reservoirs, maximizing efficiency. It is based on the second law of thermodynamics, which states that no heat engine or refrigerator can be more efficient than a Carnot cycle operating between the same temperatures. The performance of a Carnot refrigerator is determined by the temperatures of the hot and cold reservoirs, influencing the work input required to transfer heat.

Heat Transfer through Conduction

Heat transfer through conduction is the process by which thermal energy moves through a material without the material itself moving. In this scenario, the copper bar conducts heat from the hot reservoir to the cold reservoir. The rate of heat transfer can be calculated using Fourier's law, which relates the heat transfer rate to the temperature difference, the material's thermal conductivity, and the geometry of the conductor.

Work Input in Refrigeration

In refrigeration, work input is the energy supplied to the system to remove heat from the cold reservoir. The work done on the refrigerator must equal the heat absorbed from the cold reservoir minus the heat rejected to the hot reservoir, as per the first law of thermodynamics. This relationship is crucial for determining the efficiency and operational requirements of the refrigeration cycle, especially in a Carnot system.