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JEE Advanced 2024Paper 1
Exam
JEE Advanced
Year
2024
Paper
Paper 1
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Advanced 2024 · Paper 1 · Physics · Thermal Properties of Matter

The specific heat capacity of a substance is temperature dependent...

The specific heat capacity of a substance is temperature dependent and is given by the formula C=kTC=k T, where kk is a constant of suitable dimensions in SI units, and TT is the absolute temperature. If the heat required to raise the temperature of 1 kg1 \mathrm{~kg} of the substance from −73∘C-73^{\circ} \mathrm{C} to 27∘C27^{\circ} \mathrm{C} is nkn k, the value of nn is _____ [Given: 0 K=−273∘C0 \mathrm{~K}=-273^{\circ} \mathrm{C}.]

Correct answer

0

Step-by-step solution

Ti=−73∘C=200 K Tf=27∘C=300 K\begin{aligned} & \mathrm{T}_{\mathrm{i}}=-73^{\circ} \mathrm{C}=200 \mathrm{~K} \\ & \mathrm{~T}_{\mathrm{f}}=27^{\circ} \mathrm{C}=300 \mathrm{~K}\end{aligned}Q=∫msdT=∫1⋅kTdT=∫kTdT=K∫200300TdT=K2[ T2]200300=K2[3002−2002]Q=25000 K\begin{aligned} & \mathrm{Q}=\int \mathrm{msdT} \\ & =\int 1 \cdot \mathrm{kTdT} \\ & =\int \mathrm{kTdT}=\mathrm{K} \int_{200}^{300} \mathrm{TdT} \\ & =\frac{\mathrm{K}}{2}\left[\mathrm{~T}^2\right]_{200}^{300}=\frac{\mathrm{K}}{2}\left[300^2-200^2\right] \\ & \mathrm{Q}=25000 \mathrm{~K}\end{aligned}HenceHencen=25000$
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