Animated Solution for Physics - Thermodynamics: A reversible engine has an efficiency of $1/4$. If the temperature of the sink is reduced by $58^{\circ }\text{C}$, its efficiency becomes double. Calculate the temperature of the sink.

Unlocking Maximum Efficiency

Imagine you are an engineer tasked with extracting the maximum possible work from a heat source. You would build a Carnot Engine, a theoretical construct that represents the absolute pinnacle of thermodynamic efficiency.

In this problem, we are given a reversible engine operating between a hot source at temperature $T_1$ and a cold sink at temperature $T_2$. The efficiency of such an engine is governed by a beautifully simple yet profound equation:

It is absolutely critical to remember that in thermodynamics, temperatures must always be measured in Kelvin. The absolute scale is what gives this equation its physical meaning.

We are told that the engine initially operates with an efficiency of $1/4$. Let's plug this into our master equation:

By rearranging the terms, we can find the ratio of the sink temperature to the source temperature:

This gives us a direct relationship between the two temperatures:

This relationship is our anchor. No matter what happens to the engine later, this initial ratio defines the starting conditions.

Now, the problem introduces a change: the temperature of the sink is reduced by $58^{\circ }\text{C}$. Here is where many students stumble. Does a change in Celsius require a complex conversion to Kelvin?

No! A temperature difference of $1^{\circ }\text{C}$ is exactly equal to a temperature difference of $1\text{ K}$. Therefore, a reduction of $58^{\circ }\text{C}$ is mathematically identical to a reduction of $58\text{ K}$. Our new sink temperature is simply $(T_2-58)$.

With this cooler sink, the engine becomes more efficient. In fact, its efficiency doubles from $1/4$ to $1/2$. Let's set up the new equation:

Rearranging this gives:

We now have a system of equations. We can substitute our expression for $T_1$ from the first state into our new equation:

Cross-multiplying to clear the fractions:

Bringing the $T_2$ terms to one side:

Finally, solving for $T_2$:

A Note on the Options: You might notice that the options provided in the exam are in ${}^{\circ }\text{C}$, yet our mathematically rigorous answer is $174\text{ K}$. If we converted $174\text{ K}$ to Celsius, it would be $-99^{\circ }\text{C}$, which is not an option. This indicates a typographical error in the original exam paper where the unit was incorrectly printed as ${}^{\circ }\text{C}$ instead of $\text{K}$. However, the numerical value 174 matches perfectly, making option (a) the intended correct answer.