Question

A Carnot engine operates between 550°C and 20°C baths and produces 300 kJ of energy in each cycle. Find the change in entropy of the (a) hot bath and (b) cold bath, in each Carnot cycle?

a) (a) 0.6 kJ/K, (b) 0.3 kJ/K
b) (a) 0.3 kJ/K, (b) 0.6 kJ/K
c) (a) 0.6 kJ/K, (b) -0.3 kJ/K
d) (a) -0.3 kJ/K, (b) 0.6 kJ/K

Answer 1

The changes in entropy for both the hot and cold baths were calculated based on the heat transferred and their respective temperatures; however, the calculated values did not match the provided options, suggesting an error in the question or options.

Step-by-step explanation:

To find the change in entropy for the hot and cold baths in a Carnot engine, we can use the following relationship for reversible processes:

ΔS = Q/T

where ΔS is the change in entropy, Q is the heat transferred, and T is the temperature in kelvins (K).

First, we need to convert the given temperatures from degrees Celsius to kelvins:

• Hot bath: 550°C + 273.15 = 823.15 K
• Cold bath: 20°C + 273.15 = 293.15 K

(a) For the hot bath, the change in entropy ΔSh is:

ΔSh = -Qh/Th

The heat Qh is the energy produced by the engine, which is negative for the hot bath since it is losing heat:

ΔSh = -300 kJ / 823.15 K

ΔSh = -0.364 kJ/K

(b) For the cold bath, the change in entropy ΔSc is:

ΔSc = Qc/Tc

The heat Qc is positive for the cold bath since it is gaining heat:

ΔSc = 300 kJ / 293.15 K

ΔSc = 1.023 kJ/K

The actual change in entropy will be a negative value for the hot bath and a positive value for the cold bath because of the direction of heat transfer. Therefore, the correct answer is not explicitly given, suggesting there may be a mistake in the question or a misunderstanding of the entropy change concept.