Question

A Carnot engine whose sink temperature of 300K has efficiency of 40%. By how much should source temperature be increased so as to increase efficiency to 60%?

Answer 1

Answer on Question 73216, Physics, Molecular Physics, Thermodynamics

Question:

A carnot engine whose sink temperature of $300\,K$ has efficiency of $40\%$. By how much should source temperature be increased so as to increase efficiency to $60\%$?

Solution:

By the definition of the Carnot engine efficiency, we have:

$$\eta_{1} = 1 - \frac{T_{C}}{T_{H1}},$$

here, $\eta_{1} = 40\%$ is the Carnot engine efficiency, $T_{C} = 300\,K$ is the sink temperature, $T_{H1}$ is the source temperature.

From this formula, we can find the source temperature, $T_{H1}$:

$$\eta_{1} = 1 - \frac{T_{C}}{T_{H1}},$$ $$\frac{T_{C}}{T_{H1}} = 1 - \eta_{1},$$ $$T_{H1} = \frac{T_{C}}{1 - \eta_{1}} = \frac{300\,K}{1 - 0.4} = 500\,K.$$

Then, we increase the efficiency to $60\%$. We can apply the same formula for the Carnot engine efficiency and find the new source temperature, $T_{H2}$:

$$\eta_{2} = 1 - \frac{T_{C}}{T_{H2}},$$ $$\frac{T_{C}}{T_{H2}} = 1 - \eta_{2},$$ $$T_{H2} = \frac{T_{C}}{1 - \eta_{2}} = \frac{300\,K}{1 - 0.6} = 750\,K.$$

Finally, we can find the change in the source temperature:

$$\Delta T _ {H} = T _ {H 2} - T _ {H 1} = 750\,K - 500\,K = 250\,K.$$

Answer:

$$\Delta T _ {H} = 250\,K.$$

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