ΔS = Q/T. Second law: total entropy of universe always increases.

Entropy — What It Means and Why It Always Increases

Question

A reversible heat engine absorbs 1000 J of heat from a source at 500 K and rejects heat to a sink at 300 K. Calculate the change in entropy of (a) the source, (b) the sink, and (c) the universe.

Then explain: why does entropy always increase for any real (irreversible) process?

Solution — Step by Step

For a reversible (Carnot) engine, efficiency is $\eta = 1 - \frac{T_L}{T_H}$ .

\eta = 1 - \frac{300}{500} = 0.4

So work done = $0.4 \times 1000 = 400$ J, and heat rejected $Q_L = 1000 - 400 = 600$ J.

The source loses heat. Heat flows out, so $Q$ is negative for the source.

\Delta S_{\text{source}} = \frac{-Q_H}{T_H} = \frac{-1000}{500} = -2 \text{ J/K}

The sink gains 600 J. Heat flows in, so $Q$ is positive.

\Delta S_{\text{sink}} = \frac{Q_L}{T_L} = \frac{600}{300} = +2 \text{ J/K}

\Delta S_{\text{universe}} = \Delta S_{\text{source}} + \Delta S_{\text{sink}} = -2 + 2 = \mathbf{0}

For a reversible process, total entropy change of the universe is exactly zero. This is the theoretical limit — real engines always do worse.

Suppose the same engine is irreversible and rejects 700 J instead of 600 J (less efficient). Then:

\Delta S_{\text{universe}} = \frac{-1000}{500} + \frac{700}{300} = -2 + 2.33 = +0.33 \text{ J/K}

Any irreversibility — friction, heat loss, sudden expansion — always increases total entropy. Nature doesn’t allow the reverse.

Why This Works

The formula $\Delta S = Q/T$ captures something deep: entropy measures how spread out energy is. When heat flows from hot to cold, energy distributes itself more evenly across more microscopic states — that’s an entropy increase.

A reversible process is the idealised case where we move so slowly that the system is always in equilibrium. Every infinitesimal step can be undone. In this case, what the source loses in entropy, the sink gains exactly. Net change: zero.

Real processes have irreversibilities — temperature gradients, friction, turbulence. These generate extra entropy. The Second Law simply says this extra generation is always non-negative. $\Delta S_{\text{universe}} \geq 0$ is not an observation we made; it’s a statement about the direction time flows.

In JEE problems, the phrase “reversible process” is your signal that $\Delta S_{\text{universe}} = 0$ . The moment you see “irreversible”, expect $\Delta S_{\text{universe}} > 0$ . This distinction has been tested repeatedly — JEE Advanced 2024 Paper 2 asked students to identify which processes have $\Delta S = 0$ for the universe.

Alternative Method — Using the Clausius Inequality

Instead of computing source and sink separately, use the Clausius statement directly:

\oint \frac{dQ}{T} \leq 0

For our reversible engine cycle:

\frac{Q_H}{T_H} - \frac{Q_L}{T_L} = \frac{1000}{500} - \frac{600}{300} = 2 - 2 = 0

The cyclic integral equals zero — confirming reversibility. If the engine were irreversible (say $Q_L = 700$ J with the same $Q_H$ ):

\frac{1000}{500} - \frac{700}{300} = 2 - 2.33 = -0.33 < 0

The integral is negative, consistent with the Clausius inequality. This approach is faster in MCQs — you don’t need to compute $\Delta S$ for each reservoir separately.

Common Mistake

Forgetting the sign convention for heat. Students write $\Delta S_{\text{source}} = +Q_H/T_H = +2$ J/K because they use the magnitude of heat absorbed by the engine. The source is losing heat, so from the source’s perspective, $Q = -1000$ J. Always ask: is this body gaining or losing heat? Then assign the sign accordingly.

This sign error leads to $\Delta S_{\text{universe}} = +4$ J/K, which looks like entropy increased massively — but that’s wrong. A reversible engine should give exactly zero, and getting a large positive number should be your red flag to recheck signs.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the Clausius Inequality

Common Mistake

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