A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoi

Q: Why doesn't the number of reservoirs affect the entropy change of the body?

Entropy is a state function. The body's entropy change only depends on its initial and final temperatures, not on the path or the number of steps taken to reach the final temperature.

Q: Does the entropy of the *universe* change differently in both cases?

Yes! The entropy change of the *reservoirs* (surroundings) will be different. Using more reservoirs makes the process closer to being reversible, which generates less total entropy in the universe. But the question only asks for the body's entropy change.

Q: What if the heat capacity wasn't constant?

If heat capacity $C$ was a function of temperature, say $C(T)$, the entropy change would be $\int \frac{C(T)}{T} dT$. It would still be a state function, and the answer would still be exactly the same for both cases!

Q: Why is the answer $\ln 2$ if the temperatures are in Celsius?

To calculate $\Delta S = C \ln(T_f/T_i)$, temperatures MUST be in Kelvin. $200^\circ\text{C} / 100^\circ\text{C}$ is not 2 in Kelvin ($473/373 \approx 1.26$). The question likely had a typo and meant Kelvin, but the conceptual trick (both values must be equal) leads you to the correct option anyway!

Q: Are heat and work state functions too?

No, heat ($Q$) and work ($W$) are path functions. The amount of heat supplied and work done would absolutely depend on whether you used 2 reservoirs or 8 reservoirs.

Question

A solid body of constant heat capacity $1 	ext{ J/}^\circ	ext{C}$ is being heated by keeping it in contact with reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat.
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat.
In both the cases, body is brought from initial temperature $100^\circ	ext{C}$ to final temperature $200^\circ	ext{C}$. Entropy change of the body in the two cases respectively, is

Options:

(A) $\ln 2, 4 \ln 2$

(B) $\ln 2, \ln 2$

(C) $\ln 2, 2 \ln 2$

(D) $2 \ln 2, 8 \ln 2$

Sigma Percentile · Accepted Answer

### The Setup: Heating a Block

Imagine you have a solid block, and your goal is to heat it from an initial temperature of $100^\circ	ext{C}$ to a final temperature of $200^\circ	ext{C}$. The problem presents two distinct methods to achieve this.

In the first method, you sequentially place the block in contact with 2 heat reservoirs. In the second method, you use 8 heat reservoirs. The question asks for the change in entropy of the solid body in both cases. At first glance, it feels like a heavy calculus problem involving multiple integrals for each reservoir step. But is it?

### The Trap: Paths vs. States

This is where the exam setter is testing your conceptual clarity over your mathematical brute force. In thermodynamics, physical quantities are broadly classified into two categories: **Path Functions** and **State Functions**.

Path functions, like Heat ($Q$) and Work ($W$), depend entirely on *how* a process is carried out. If the question had asked for the total heat supplied or the entropy change of the *universe* (which includes the reservoirs), the answers would be wildly different for the 2-reservoir case versus the 8-reservoir case.

### The Golden Rule: Entropy is a State Function

However, the question specifically asks for the entropy change of the **body** (the system). Entropy ($S$), just like Internal Energy ($U$), is a **State Function**.

This means the change in entropy $\Delta S$ depends *only* on the initial state and the final state of the system. It is completely blind to the path taken to get there. Whether you heat the block using 2 reservoirs, 8 reservoirs, or an infinite number of reservoirs, the entropy change of the block is mathematically locked to its initial and final temperatures:

$$\Delta S = \int_{T_i}^{T_f} \frac{dQ_{	ext{rev}}}{T} = \int_{T_i}^{T_f} \frac{C dT}{T} = C \ln\left(\frac{T_f}{T_i}ight)$$

Since $T_i$ and $T_f$ are identical in both cases, the entropy change $\Delta S$ must be exactly the same!

### The Exam Hack: Spotting the Answer

Armed with this profound conceptual truth, let's look at the options provided:
(a) $\ln 2, 4 \ln 2$
(b) $\ln 2, \ln 2$
(c) $\ln 2, 2 \ln 2$
(d) $2 \ln 2, 8 \ln 2$

We know that $\Delta S_{	ext{case 1}} = \Delta S_{	ext{case 2}}$. The only option where both values are identical is **Option (b)**.

*Ninja Note:* If you actually tried to calculate $C \ln(T_f/T_i)$, you must use Kelvin. $T_i = 373 	ext{ K}$ and $T_f = 473 	ext{ K}$, which does not yield $\ln 2$. The exam setter likely made a typo by treating Celsius as an absolute scale (where $200/100 = 2$). But because you understood the concept of state functions, you bypassed the calculation entirely and dodged the typo trap!

Select Answer:

Visualized Solution

$T_{i} \to T_{f}$

\text{Different Paths}

\text{State Function}

\Delta S_1 = \Delta S_2

\text{Final Answer}

The Sigma Insight: Second Law of Thermodynamics

The Setup

The Trap

The Golden Rule

The Exam Hack

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