A current carrying wire heats a metal rod. The wire provides a constant power ( P ) to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature ( T ) in the metal rod changes with time ( t ) as T ( t )= T 0 (1+β t 1 / 4 ) where β is a constant with appropriate dimension while T 0 is a constant with dimension of temperature. The heat capacity of the metal is

frac 4 P ( T ( t )- T 0 ) 3 β 4 T 0 4

frac 4 P ( T ( t )- T 0 ) 4 β 4 T 0 5

frac 4 P ( T ( t )- T 0 ) β 4 T 0 2

frac 4 P ( T ( t )- T 0 ) 2 β 4 T 0 2

Physics · Thermodynamics

JEE Advanced 2019 — Paper 1 — Question 3

A current carrying wire heats a metal rod. The wire provides a constant power ( P ) to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature $(\mathrm{T})$ in the metal rod changes with time ( t ) as $\mathrm{T}(\mathrm{t})=\mathrm{T}_{0}\left(1+\beta \mathrm{t}^{1 / 4}\right)$ where $\beta$ is a constant with appropriate dimension while $\mathrm{T}_{0}$ is a constant with dimension of temperature. The heat capacity of the metal is

Option A:
$\frac{4 \mathrm{P}\left(\mathrm{T}(\mathrm{t})-\mathrm{T}_{0}\right)^{4}}{\beta^{4} \mathrm{~T}_{0}^{5}}$
Option B:
$\frac{4 \mathrm{P}\left(\mathrm{T}(\mathrm{t})-\mathrm{T}_{0}\right)}{\beta^{4} \mathrm{~T}_{0}^{2}}$
Option C:
$\frac{4 \mathrm{P}\left(\mathrm{T}(\mathrm{t})-\mathrm{T}_{0}\right)^{2}}{\beta^{4} \mathrm{~T}_{0}^{2}}$
Option D:
$\frac{4 \mathrm{P}\left(\mathrm{T}(\mathrm{t})-\mathrm{T}_{0}\right)^{3}}{\beta^{4} \mathrm{~T}_{0}^{4}}$
Correct

Answer: D

Step-by-step solution

At equilibrium, $C \frac{d T}{d t}=P$

$\frac{\mathrm{dT}}{\mathrm{dt}}=\frac{\mathrm{T}_{0} \beta}{4} \mathrm{t}^{-\frac{3}{4}}$

So heat capacity $C=\frac{4 P}{\beta T_{0}} t^{\frac{3}{4}}$

From the given equation $\frac{T(t)-T_{0}}{\beta T_{0}}=t^{\frac{1}{4}}$

So $\mathrm{t}^{\frac{3}{4}}=\frac{\left(\mathrm{T}(\mathrm{t})-\mathrm{T}_{0}\right)^{3}}{\beta^{3} \mathrm{~T}_{0}^{3}}$

So $\mathrm{C}=\frac{4 \mathrm{P}}{\beta^{4} \mathrm{~T}_{0}^{4}}\left(\mathrm{~T}(\mathrm{t})-\mathrm{T}_{0}\right)^{3}$

Answer key and solution verified before publishing.

Practise Thermodynamics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Advanced 2019
Paper: Paper 1
Subject: Physics
Chapter: Thermodynamics
Topic: First Law of Thermodynamics

One mole of a monatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature ( V- T) diagram. The…

Answer: D

Step-by-step solution

Practise Thermodynamics

More Thermodynamics questions from this paper