JEE Advance - Physics (2024 - Paper 1 Online - No. 8)

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: $0 \mathrm{~K}=-273{ }^{\circ} \mathrm{C}$.]

Answer
25000

Explanation

To solve this problem, we need to integrate the heat capacity over the given temperature range because the specific heat capacity is temperature dependent. We are given that the specific heat capacity $$C$$ is defined as $$C = kT$$, where $$k$$ is a constant, and $$T$$ is the absolute temperature.

The heat required to raise the temperature, $$Q$$, can be found using the following integral:

$$ Q = \int_{T_1}^{T_2} C \, dT $$

Given $$C = kT$$, the integral becomes:

$$ Q = \int_{T_1}^{T_2} kT \, dT $$

We need to convert the given temperatures from Celsius to Kelvin. The temperatures given are:

  • Initial temperature: $$-73^{\circ} \mathrm{C}$$
  • Final temperature: $$27^{\circ} \mathrm{C}$$

In Kelvin, these temperatures are:

  • $$T_1 = -73^{\circ} \mathrm{C} + 273 = 200 \, \mathrm{K}$$
  • $$T_2 = 27^{\circ} \mathrm{C} + 273 = 300 \, \mathrm{K}$$

Now we can evaluate the integral:

$$ Q = k \int_{200}^{300} T \, dT $$

Integrating, we get:

$$ Q = k \left[ \frac{T^2}{2} \right]_{200}^{300} $$

Substituting the limits of integration:

$$ Q = k \left[ \frac{300^2}{2} - \frac{200^2}{2} \right] $$

Solving the values inside the brackets:

$$ Q = k \left[ \frac{90000}{2} - \frac{40000}{2} \right] $$

$$ Q = k \left[ 45000 - 20000 \right] $$

$$ Q = k \cdot 25000 $$

We are given that this heat is equal to $$nk$$:

$$ nk = k \cdot 25000 $$

Dividing both sides by $$k$$, we get:

$$ n = 25000 $$

Thus, the value of $$n$$ is 25000.

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