Animated Solution for Physics - Thermodynamics: A Carnot engine having an efficiency of $\frac{1}{10}$ is being used as a refrigerator. If the work done on the refrigerator is $10\text{ J}$, then the amount of heat absorbed from the reservoir at lower temperature is:

Imagine a refrigerator operating in your kitchen. It performs a seemingly magical task: it extracts heat from a cold region (the inside of the fridge) and dumps it into a hotter region (your kitchen). However, the Second Law of Thermodynamics tells us that heat cannot flow from a colder body to a hotter body on its own. It requires external effort. This is why your refrigerator needs to be plugged into an electrical outlet to do work on the system.

In this problem, we are dealing with an idealized reversible machine—a Carnot engine—that is being run in reverse to act as a refrigerator. We are given its efficiency $\eta$ when it operates as an engine, and we need to find out how much heat it absorbs from the cold reservoir when a specific amount of work is done on it.

The performance of a refrigerator is measured not by 'efficiency', but by its Coefficient of Performance (COP), denoted by $\beta$. The COP tells us how much cooling we get for every unit of work we put in.

There is a beautiful and direct mathematical relationship between the efficiency $\eta$ of a Carnot engine and the COP $\beta$ of the same machine run as a Carnot refrigerator:

We are given that the efficiency of the engine is $\eta =\frac{1}{10}$. Let's substitute this value into our master equation to find the COP.

Simplifying the numerator, we get $\frac{9}{10}$. Dividing this by $\frac{1}{10}$, the denominators cancel out perfectly:

This means that for every $1\text{ J}$ of work put into the refrigerator, it extracts $9\text{ J}$ of heat from the cold reservoir. That is a highly effective cooling machine!

Now, let's return to the fundamental definition of the Coefficient of Performance. Physically, it is the ratio of the desired output (heat extracted, $Q_2$) to the required input (work done, $W$):

We have calculated $\beta =9$, and the problem states that the work done on the refrigerator is $W=10\text{ J}$. Substituting these values into the definition gives:

Multiplying both sides by $10$, we arrive at our final answer:

The refrigerator absorbs $90\text{ J}$ of heat from the cold reservoir.

As a thought experiment, what if we wanted to know the total heat rejected to the hot room? By the First Law of Thermodynamics (conservation of energy), the heat rejected $Q_1$ must equal the heat absorbed plus the work done: $Q_1=Q_2+W=90+10=100\text{ J}$.