Animated Solution for Physics - Thermodynamics: A heat engine has an efficiency of $1/6$. When the temperature of sink is reduced by $62^{\circ }\text{C}$, its efficiency get doubled. The temperature of the source is

Imagine a classic heat engine. It operates by extracting heat energy ($Q_1$) from a hot reservoir called the source at temperature $T_1$. It converts a portion of this heat into useful mechanical work ($W$) and rejects the remaining, unused heat ($Q_2$) into a cold reservoir called the sink at temperature $T_2$.

The efficiency ($\eta$) of this engine tells us how good it is at converting heat into work. For an ideal reversible engine, this efficiency depends entirely on the absolute temperatures of the source and the sink, given by the beautiful relation:

Initially, we are told that the engine has an efficiency of $1/6$. This means that for every 6 Joules of heat it takes from the source, it only manages to do 1 Joule of work. Let's plug this into our efficiency formula:

By rearranging this equation, we can isolate the ratio of the sink temperature to the source temperature:

This ratio, $\frac{5}{6}$, is the foundational key to unlocking the rest of the problem. We will hold onto it for the next step.

Next, the problem introduces a change: the temperature of the sink is reduced by $62^{\circ }\text{C}$. A crucial concept in thermodynamics is that a change in temperature ($\Delta T$) is numerically identical whether you measure it in Celsius or Kelvin. Therefore, a drop of $62^{\circ }\text{C}$ is exactly the same as a drop of $62\text{ K}$.

Our new sink temperature becomes $(T_2-62)$. Because the sink is now colder, the temperature gradient is steeper, and the engine becomes more efficient. The problem states the efficiency doubles, becoming $2\times \frac{1}{6}=\frac{1}{3}$. Let's set up our second equation:

Rearranging this gives us:

Now, we have a fraction on the left side that we can split into two parts. This is a strategic algebraic move:

Look closely at the first term, $\frac{T_2}{T_1}$. We already found its value in our initial state! It is exactly $\frac{5}{6}$. By substituting this value, we completely eliminate the unknown $T_2$ from our equation:

Now, it is a straightforward path to find $T_1$. Let's isolate the term containing $T_1$:

Finding a common denominator (which is 6), we get:

Cross-multiplying yields the absolute temperature of the source:

Kelvin vs. Celsius

We have successfully found the source temperature to be $372\text{ K}$. However, if you rush to the options, you might be confused or tempted to pick a wrong answer. The options are provided in degrees Celsius!

Never forget to check your units at the final step. To convert from Kelvin back to Celsius, we must subtract 273:

And there we have it. The temperature of the source is $99^{\circ }\text{C}$, which perfectly matches option (d).