Animated Solution for Physics - Thermodynamics: If minimum possible work is done by a refrigerator in converting $100\text{ g}$ of water at $0^{\circ }\text{C}$ to ice, how much heat (in cal) is released to the surroundings at temperature $27^{\circ }\text{C}$ to the nearest integer …… ? (Take, latent heat of ice $=80\text{ cal/g}$)

Have you ever wondered how a refrigerator actually works? It feels almost like magic—taking heat from a cold place (like the water you want to freeze) and dumping it into a warmer place (like your kitchen). But as thermodynamics tells us, this magic comes at a cost: Work.

In this fascinating problem, we are tasked with freezing $100\text{ g}$ of water at $0^{\circ }\text{C}$ into ice, while the surrounding room is at a balmy $27^{\circ }\text{C}$. The question asks for the heat released to the surroundings, given a very specific and crucial condition: the refrigerator does the minimum possible work.

Let's visualize the system. A refrigerator operates between two thermal reservoirs: 1. The Cold Reservoir: This is the water at $0^{\circ }\text{C}$. We will call its temperature $T_1$. 2. The Hot Reservoir: This is the surrounding room at $27^{\circ }\text{C}$. We will call its temperature $T_2$.

The refrigerator's job is to extract a certain amount of heat, let's call it $Q_1$, from the cold reservoir. To do this, an external compressor must do work, $W$, on the system. Finally, the refrigerator dumps the total energy—both the extracted heat and the work done—into the hot reservoir as rejected heat, $Q_2$.

First, we need to figure out exactly how much heat must be extracted to freeze the water. Since the water is already at $0^{\circ }\text{C}$ and we are turning it into ice at $0^{\circ }\text{C}$, there is no change in temperature. We only need to account for the phase change using the latent heat of fusion.

Given the mass $m=100\text{ g}$ and the latent heat $L=80\text{ cal/g}$, we can easily compute this:

So, the refrigerator must pull $8000\text{ cal}$ of heat out of the water.

Now, let's decode the phrase "minimum possible work". In thermodynamics, a process requires the minimum amount of work if and only if it is completely reversible. A reversible refrigerator is an ideal Carnot refrigerator.

For a Carnot cycle, the ratio of the heat exchanged with a reservoir to the absolute temperature of that reservoir is constant. This gives us our master equation:

Before we plug in the numbers, we must avoid the most common trap in thermodynamics: Temperatures must always be in Kelvin!

Let's convert our temperatures: - $T_1=0^{\circ }\text{C}+273=273\text{ K}$ - $T_2=27^{\circ }\text{C}+273=300\text{ K}$

Now, we substitute these absolute temperatures and our known $Q_1$ into the Carnot relation:

All that remains is the arithmetic. We isolate $Q_2$ to find the heat rejected to the surroundings:

The question asks for the answer to the nearest integer. Rounding $8791.2$ gives us our final answer:

What if we wanted to know exactly how much work the compressor did? By the First Law of Thermodynamics (Conservation of Energy), the heat rejected is the sum of the heat absorbed and the work done:

This means the refrigerator required $791\text{ cal}$ of electrical work to move $8000\text{ cal}$ of heat. This ratio is known as the Coefficient of Performance (COP), which in this ideal case is $\beta =\frac{8000}{791}\approx 10.1$. Real-world refrigerators have much lower COPs due to friction and irreversible losses, but the Carnot limit shows us the absolute best we can ever hope to achieve!