Question:

A refrigerator operates between a cold reservoir at a temperature of 5°C and a hot reservoir at a temperature of 30°C. During the cooling cycle, 300 J of heat is absorbed from the cold reservoir. Calculate the work done by the refrigerator during this cycle.

Answer:

To calculate the work done by the refrigerator during this cycle, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

Mathematically, this can be expressed as:

ΔU = Q - W

Where: ΔU is the change in internal energy of the system, Q is the heat added to the system, and W is the work done by the system.

Since the refrigerator is operating in a cooling cycle, the heat added to the system is negative (-300 J). We can substitute the known values into the equation to solve for W:

ΔU = -300 J - W

We also know that the change in internal energy of the system is equal to zero since the refrigerator returns to its initial state after completing a cooling cycle. Therefore:

ΔU = 0

Substituting this into the equation:

0 = -300 J - W

Rearranging the equation to solve for W:

W = -300 J

Therefore, the work done by the refrigerator during the cooling cycle is -300 J.

Note: The negative sign indicates that work is done on the refrigerator, as it is an input of energy to the system.