JEE Advanced 2024 — Paper 1 — PHY-III solution

JEE Advanced2024Paper 1PHY-III

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 $1$ kg of the substance from $- 7 3^{\circ}$ C to $2 7^{\circ}$ C is $nk$ , the value of $n$ is ______.

[Given: $0$ K $= - 27 3^{\circ}$ C.]

Your answer

Solution

Convert temperatures: $T_{1} = 200$ K and $T_{2} = 300$ K.

For mass $m$ with $C = k T$ ,

$Q = \int_{T_{1}}^{T_{2}} m C d T = m \int_{200}^{300} k T d T = m \cdot \frac{k}{2} (T_{2}^{2} - T_{1}^{2})$

With $m = 1$ kg:

$Q = \frac{k}{2} (30 0^{2} - 20 0^{2}) = \frac{k}{2} (90000 - 40000) = \frac{k}{2} \cdot 50000 = 25000 k$ .

$∴ n = 25000$ .

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