Question: A heat engine operates between a hot reservoir at a temperature of 600°C and a cold reservoir at a temperature of 300°C. The engine emits waste heat to the cold reservoir. The engine has an efficiency of 40%. Calculate the entropy change of the hot reservoir and the cold reservoir during one cycle of the engine.

Solution:

First, let's define the efficiency of a heat engine. The efficiency (η) of a heat engine is given by the equation:

η = 1 - (Tc/Th)

where Th is the temperature of the hot reservoir and Tc is the temperature of the cold reservoir.

Given that the efficiency of the heat engine is 40%, we can substitute the values into the equation to find the Tc/Th ratio.

0.40 = 1 - (Tc/600)

Rearranging the equation, we have:

Tc/600 = 1 - 0.40 Tc/600 = 0.60

To find the value of Tc, we can multiply both sides of the equation by 600:

Tc = 0.60 x 600 Tc = 360°C

Now that we know the temperatures of both the hot reservoir (Th = 600°C) and the cold reservoir (Tc = 360°C), we can calculate the entropy change using the equation:

ΔS = Q / T

where Q is the heat transferred and T is the temperature.

For the hot reservoir, the entropy change can be calculated as:

ΔS_hot = Q_hot / Th

Since the heat transferred to the hot reservoir is equal to the heat absorbed by the engine (Q_hot) and the temperature remains constant throughout the process, the equation becomes:

ΔS_hot = Q_hot / Th

For the cold reservoir, the entropy change can be calculated as:

ΔS_cold = Q_cold / Tc

Since the heat transferred to the cold reservoir is equal to the waste heat emitted by the engine (Q_cold) and the temperature remains constant throughout the process, the equation becomes:

ΔS_cold = Q_cold / Tc

However, we need to consider that the heat absorbed by the engine (Q_hot) is equal to the ideal work done by the engine (W) plus the waste heat emitted by the engine (Q_cold). Therefore, we can re-write the equation for the entropy change of the hot reservoir:

ΔS_hot = (W + Q_cold) / Th

Using the first law of thermodynamics, we know that the work done by the engine (W) is equal to the heat absorbed by the engine (Q_hot) minus the heat emitted by the engine (Q_cold):

W = Q_hot - Q_cold

Substituting this into the equation for ΔS_hot, we get:

ΔS_hot = (Q_hot - Q_cold + Q_cold) / Th ΔS_hot = Q_hot / Th

Now, let's substitute the values into the equations:

ΔS_hot = Q_hot / Th = Q_hot / 600

ΔS_cold = Q_cold / Tc = Q_cold / 360

From the efficiency equation, we know that the heat absorbed by the engine (Q_hot) is equal to 0.40 times the heat input to the engine (Q_in):

Q_hot = 0.40 x Q_in

Since the efficiency is given as 40%, the ratio of Q_hot to Q_in is 0.40. Therefore, we can write:

Q_hot = 0.40 x Q_in

Since Q_hot is equal to the heat absorbed by the engine, we can also say:

Q_in = Q_hot

Substituting this into the equation for ΔS_hot, we get:

ΔS_hot = Q_in / 600

For the cold reservoir, we know that the heat emitted by the engine (Q_cold) is equal to 0.60 times the heat input to the engine (Q_in):

Q_cold = 0.60 x Q_in

Substituting this into the equation for ΔS_cold, we get:

ΔS_cold = (0.60 x Q_in) / 360

Now we have expressions for the entropy change of the hot reservoir and the cold reservoir. Let's substitute the values and calculate them.

ΔS_hot = Q_in / 600

ΔS_cold = (0.60 x Q_in) / 360

The entropy change for the hot reservoir, ΔS_hot, is equal to Q_in divided by 600. However, the value of Q_in is not given in the problem statement, so we cannot determine the exact value of ΔS_hot.

Similarly, the entropy change for the cold reservoir, ΔS_cold, is equal to 0.60 times Q_in divided by 360. Since we do not know the value of Q_in, we cannot determine the exact value of ΔS_cold.

Therefore, without knowing the value of Q_in, we cannot calculate the exact entropy changes of the hot and cold reservoirs during one cycle of the engine.