Physics · Thermodynamics

JEE Advanced 2024 — Paper 1 — Question 25

The specific heat capacity of a substance is temperature dependent and is given by the formula C $=\mathrm{kT}$ , where k is a constant of suitable dimensions in SI units,and T is the absolute temperature. If the heat required to raise the temperature of 1 kg of the substance from $-73^{\circ} \mathrm{C}$ to $27^{\circ} \mathrm{C}$ is nk , the value of $n$ is $\_\_\_\_\_$ . [Given: $0 \mathrm{~K}=-273^{\circ} \mathrm{C}$ ]

Answer: 25000

Numerical answer — enter this value.

Step-by-step solution

$\quad d Q=m S d T$

$\int d Q=\int 1 . K T d T$

$\Delta Q=K \int_{200}^{300} T d T$

$\Delta \mathrm{Q}=\frac{\mathrm{K}}{2}\left[\mathrm{~T}^{2}\right]_{200}^{300}$

$\Delta Q=\frac{K}{2} \times(9-4) \times 10^{4}$

$\Delta Q=25000 K$

$\mathrm{n}=25000$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2024
Paper: Paper 1
Subject: Physics
Chapter: Thermodynamics
Topic: Calculation of Work, Heat and Internal Energy

One mole of a monatomic ideal gas undergoes the cyclic process J arrow K arrow L arrow M arrow J , as shown in the P-T diagram. Match the…

Answer: 25000

Step-by-step solution

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