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Question #42182

Difficulty: 8/10 | 🟢 Free

The Question

A uniform metal rod of mass $m$ , length $L$ , cross-sectional area $A$ , specific heat capacity $c$ (assumed constant), and Young's modulus $Y$ is rigidly fixed between two unyielding walls at an absolute temperature $T_0$ . The coefficient of linear expansion of the rod varies with temperature as $\alpha = \beta T$ , where $\beta$ is a positive constant. The rod is heated to a temperature $2T_0$ . Assuming the walls do not move and the rod does not buckle, what is the total heat $\Delta Q$ supplied to the rod?

Options

A.
$m c T_0 + \frac{9}{8} Y A L \beta^2 T_0^4$
B.
$m c T_0 + \frac{9}{4} Y A L \beta^2 T_0^4$
C.
$m c T_0$
D.
$m c T_0 - \frac{9}{8} Y A L \beta^2 T_0^4$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

First Law Application|Thermal Stress in Constrained Rods|Calculus (Integration)

Why This Question Is Tricky

Common Mistakes

This question tests the application of the First Law of Thermodynamics to a constrained solid undergoing thermal stress.

The primary trap is the misconception that the rod does macroscopic work against the walls ( $W = \int F dx$ ), leading students to erroneously add a work term (often equal to the elastic energy) to the heat supplied.

In reality, because the walls are unyielding, the boundary displacement is zero, meaning $W = 0$ .

The heat supplied goes entirely into increasing the internal energy, which comprises both the standard thermal energy ( $mc\Delta T$ ) and the macroscopic elastic potential energy stored due to the constrained thermal expansion.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Thermodynamics - Class 11 Part 2
Concept	First Law Application
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free