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Question #42506

Difficulty: 8/10 | 🟢 Free

The Question

Two identical solid spheres, A and B, each of mass $m$ and constant specific heat capacity $c$ , are initially at temperatures $T_A$ and $T_B$ respectively, with $T_A > T_B > T_0$ . They are placed inside a large evacuated enclosure maintained at a constant temperature $T_0$ . The spheres exchange heat with each other via a conducting wire, and simultaneously lose heat to the enclosure exclusively via thermal radiation according to the Stefan-Boltzmann law, $\frac{dQ}{dt} = -\epsilon \sigma A (T^4 - T_0^4)$ . After a long time, both spheres reach thermal equilibrium with the enclosure. What is the total change in entropy of the universe ( $\Delta S_{\text{univ}}$ ) during this entire process?

Options

A.
$mc \ln\left(\frac{T_0^2}{T_A T_B}\right) + \frac{mc(T_A + T_B - 2T_0)}{T_0}$
B.
$mc \ln\left(\frac{T_0^2}{T_A T_B}\right) + \frac{\epsilon \sigma A}{T_0} (T_A^4 + T_B^4 - 2T_0^4)$
C.
$mc \ln\left(\frac{(T_A + T_B)^2}{4 T_A T_B}\right) + \frac{mc(T_A + T_B - 2T_0)}{T_0}$
D.
$mc \ln\left(\frac{T_0^2}{T_A T_B}\right)$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Entropy and Reversibility|Calorimetry with Heat Loss|Newton’s Law of Cooling Limits|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's fundamental understanding of entropy as a state function and the First Law of Thermodynamics, specifically trapping those who get bogged down by complex, non-linear rate equations.

The presence of simultaneous conduction and Stefan-Boltzmann radiative cooling ( $\propto T^4$ ) acts as a massive distractor, tempting students to set up unsolvable coupled differential equations to integrate $dS = dQ/T$ over time.

The trap exploits the misconception that a complex, path-dependent heat transfer mechanism alters the total entropy change.

By recognizing that the enclosure acts as an infinite thermal reservoir at constant $T_0$ , students can bypass the rate equations entirely: the system's entropy change depends only on the initial and final temperatures, and the surroundings' entropy change depends only on the total energy conserved and transferred, rendering the non-linear cooling path mathematically irrelevant.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Thermodynamics - Class 11 Part 2
Concept	Entropy and Reversibility
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free