JEE Advanced · Physics · 12. Thermal Properties
The specific heat capacity of a substance is temperature dependent and is given by the formula \(C=k T\), 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 \(1 \mathrm{~kg}\) of the substance from \(-73^{\circ} \mathrm{C}\) to \(27^{\circ} \mathrm{C}\) is \(n k\), the value of \(n\) is _____
[Given: \(0 \mathrm{~K}=-273^{\circ} \mathrm{C}\).]
- A 25000
- B 35000
- C 45000
- D 55000
Answer & Solution
Correct Answer
(A) 25000
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{T}_{\mathrm{i}}=-73^{\circ} \mathrm{C}=200 \mathrm{~K} \\ & \mathrm{~T}_{\mathrm{f}}=27^{\circ} \mathrm{C}=300 \mathrm{~K}\end{aligned}\)
\(\begin{aligned} & \mathrm{Q}=\int \mathrm{msdT} \\ & =\int 1 \cdot \mathrm{kTdT} \\ & =\int \mathrm{kTdT}=\mathrm{K} \int_{200}^{300} \mathrm{TdT} \\ & =\frac{\mathrm{K}}{2}\left[\mathrm{~T}^2\right]_{200}^{300}=\frac{\mathrm{K}}{2}\left[300^2-200^2\right] \\ & \mathrm{Q}=25000 \mathrm{~K}\end{aligned}\)
Hence \(n=25000\)
\(\begin{aligned} & \mathrm{Q}=\int \mathrm{msdT} \\ & =\int 1 \cdot \mathrm{kTdT} \\ & =\int \mathrm{kTdT}=\mathrm{K} \int_{200}^{300} \mathrm{TdT} \\ & =\frac{\mathrm{K}}{2}\left[\mathrm{~T}^2\right]_{200}^{300}=\frac{\mathrm{K}}{2}\left[300^2-200^2\right] \\ & \mathrm{Q}=25000 \mathrm{~K}\end{aligned}\)
Hence \(n=25000\)
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