Animated Solution for Physics - Thermodynamics: A Carnot engine operates between two reservoirs of temperatures 900 K and 300 K. The engine performs 1200 J of work per cycle. The heat energy (in J) delivered by the engine to the low temperature reservoir in a cycle, is ......... .

The Carnot engine is the crown jewel of classical thermodynamics. Conceived by Sadi Carnot in 1824, it represents the theoretical limit of how efficiently we can convert heat into useful work.

Imagine a thermal water wheel. Just as a water wheel extracts energy from water falling from a high elevation to a low elevation, a heat engine extracts work as heat "falls" from a high-temperature source to a low-temperature sink.

In this problem, we are tasked with finding exactly how much heat is dumped into the cold sink during each cycle of such an engine.

Let's break down the physical parameters given to us. We have a hot reservoir, which acts as our thermal source.

The temperature of this source is $T_1=900\text{ K}$.

On the other end, we have a cold reservoir, our thermal sink, maintained at a temperature of $T_2=300\text{ K}$.

Between these two reservoirs operates our Carnot engine. During each complete cycle, the engine performs a net mechanical work of $W=1200\text{ J}$.

Our goal is to determine $Q_2$, the heat energy delivered to the low-temperature reservoir.

The defining characteristic of a perfectly reversible Carnot engine is the elegant relationship between the heat transferred and the absolute temperatures of the reservoirs.

Because the total entropy change for a reversible cycle is zero, the ratio of the heat absorbed from the source ($Q_1$) to the heat rejected to the sink ($Q_2$) is exactly equal to the ratio of their respective absolute temperatures.

Mathematically, this is expressed as:

This equation is our primary tool. It tells us that the thermodynamics of the engine are entirely dictated by the temperatures it operates between.

Now, we can substitute our known temperatures into this master equation.

Simplifying this fraction is straightforward. Nine hundred divided by three hundred gives us exactly three.

This is a profound physical insight. It means that for every joule of heat the engine rejects to the cold sink, it must have absorbed three joules of heat from the hot source.

We can rearrange this to express the absorbed heat in terms of the rejected heat:

Next, we must invoke the First Law of Thermodynamics, which is the principle of conservation of energy.

For a cyclic process, the internal energy of the engine returns to its initial state at the end of each cycle. Therefore, the net work done by the engine must equal the net heat added to the system.

The engine takes in heat $Q_1$ and throws away heat $Q_2$. The difference between these two quantities is the useful work $W$ extracted by the engine.

We are given that the work done is $1200\text{ J}$. We also know from our previous step that $Q_1=3Q_2$.

Let's substitute these values into our energy conservation equation.

This simplifies to a very simple linear equation:

Dividing both sides by two, we isolate $Q_2$:

And there we have it! The engine delivers exactly $600\text{ J}$ of heat energy to the low-temperature reservoir in each cycle.

This problem beautifully illustrates the interplay between the First and Second Laws of Thermodynamics. The First Law ensures energy is conserved ($Q_1=W+Q_2$), while the Second Law (via Carnot's principle) dictates the exact proportions of how that energy must be split based on the operating temperatures.