Animated Solution for Physics - Thermodynamics: A Carnot engine operating between temperatures $T_1$ and $T_2$ has efficiency $\frac{1}{6}$. When $T_2$ is lowered by $62\text{ K}$, its efficiency increases to $1/3$. Then, $T_1$ and $T_2$ are respectively

Imagine a perfectly ideal machine, a theoretical construct that extracts heat from a blazing hot source, converts a portion of it into useful mechanical work, and dumps the remaining heat into a cold sink. This is the Carnot Engine, the gold standard of thermodynamics.

The beauty of the Carnot engine lies in its simplicity. Its efficiency, denoted by $\eta$, doesn't depend on the working substance or the mechanical design. It depends exclusively on the absolute temperatures of the hot source ($T_1$) and the cold sink ($T_2$). The master equation governing this is:

This elegant formula tells us a profound truth: to make an engine more efficient, you must either make the source hotter (increase $T_1$) or make the sink colder (decrease $T_2$). Let's see this principle in action through our problem.

We are told that initially, the Carnot engine operates with an efficiency of $\frac{1}{6}$. Let's plug this into our master equation:

By rearranging the terms, we can isolate the ratio of the two temperatures. Moving $\frac{T_2}{T_1}$ to the left and $\frac{1}{6}$ to the right gives us:

This is a crucial piece of information. It tells us that the sink temperature is exactly $\frac{5}{6}$ of the source temperature. Let's hold onto this ratio; it will be our key to unlocking the final answer.

Now, the problem introduces a twist. The temperature of the sink is lowered by $62\text{ K}$. So, our new sink temperature is $(T_2-62)$.

As thermodynamics dictates, a colder sink means a larger temperature gradient, which translates to higher efficiency. The problem states the new efficiency jumps to $\frac{1}{3}$. Let's set up the equation for this new, more efficient state:

Just like before, let's isolate the temperature term:

We now have a slightly complex fraction on the left side. The trick here is to split the numerator:

Do you see it? The term $\frac{T_2}{T_1}$ has reappeared! We already know from our initial analysis that $\frac{T_2}{T_1}=\frac{5}{6}$. By substituting this value, we eliminate $T_2$ entirely, leaving us with a straightforward equation for $T_1$:

Let's solve for $T_1$. We move $\frac{62}{T_1}$ to one side and bring the fractions together:

To subtract these fractions, we need a common denominator. $\frac{2}{3}$ is equivalent to $\frac{4}{6}$:

Cross-multiplying gives us the absolute temperature of the hot source:

With $T_1$ secured, finding $T_2$ is a walk in the park. We return to our trusty ratio from the first step:

Substituting the value of $T_1$ we just found:

And there we have it! The source temperature is $372\text{ K}$ and the sink temperature is $310\text{ K}$. This problem beautifully illustrates how manipulating the thermal reservoirs directly impacts the performance of a heat engine.