Question:

A heat engine operates between two reservoirs of temperature 500 K and 300 K. The engine absorbs 5000 J of heat from the hot reservoir and 3500 J of work is done by the system.

a) Calculate the heat rejected by the engine to the cold reservoir.

b) Determine the efficiency of the engine.

c) Calculate the change in entropy of the engine.

Answer:

a) To find the heat rejected by the engine to the cold reservoir, we can use the first law of thermodynamics, which states that the energy input is equal to the energy output plus the energy transferred as heat:

Energy input = Energy output + Heat transferred

Given that the engine absorbs 5000 J of heat from the hot reservoir (energy input) and 3500 J of work is done by the system (energy output), we can rearrange the equation to solve for the heat transferred:

Heat transferred = Energy input - Energy output

Heat transferred = 5000 J - 3500 J

Heat transferred = 1500 J

Therefore, the heat rejected by the engine to the cold reservoir is 1500 J.

b) The efficiency of a heat engine is defined as the ratio of the work done by the engine to the energy input:

Efficiency = Work done / Energy input

Given that 3500 J of work is done by the system and the energy input is 5000 J, we can substitute these values into the equation to calculate the efficiency:

Efficiency = 3500 J / 5000 J

Efficiency = 0.7 or 70%

Therefore, the efficiency of the engine is 70%.

c) The change in entropy of the engine can be calculated using the equation:

$\Delta S = \frac{{Q}}{{T}}$

where ΔS is the change in entropy, Q is the heat transferred, and T is the temperature in Kelvin.

Given that Q is 1500 J (heat transferred) and the temperature of the cold reservoir is 300 K, we can substitute these values into the equation to calculate the change in entropy:

$\Delta S = \frac{{1500 J}}{{300 K}}$

$\Delta S = 5 J/K$

Therefore, the change in entropy of the engine is 5 J/K.