JEE Advanced 2024 Paper 1 · Q08 · Calorimetry
The specific heat capacity of a substance is temperature dependent and is given by the formula $C = \kappa T$, where $\kappa$ is a constant of suitable dimensions in SI units, and $T$ is the absolute temperature. If the heat required to raise the temperature of $1\,\text{kg}$ of the substance from $-73\,^\circ\text{C}$ to $27\,^\circ\text{C}$ is $n\kappa$, the value of $n$ is _____.
[Given: $0\,\text{K} = -273\,^\circ\text{C}$.]
Reveal answer + step-by-step solution
Correct answer:25000
Solution
Convert temperatures to Kelvin: $T_1 = -73 + 273 = 200\,\text{K}$, $T_2 = 27 + 273 = 300\,\text{K}$. For mass $m = 1\,\text{kg}$, heat required is $Q = \int_{T_1}^{T_2} m\,C\,dT = m\kappa\int_{200}^{300} T\,dT = m\kappa\,\dfrac{T^2}{2}\bigg|_{200}^{300} = \dfrac{m\kappa}{2}(300^2 - 200^2) = \dfrac{1\cdot\kappa}{2}(90000 - 40000) = \dfrac{\kappa}{2}\cdot 50000 = 25000\,\kappa$. Therefore $n = 25000$.
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