JEE MAIN - Physics (2023 - 6th April Evening Shift - No. 12)

A body cools in 7 minutes from $$60^{\circ} \mathrm{C}$$ to $$40^{\circ} \mathrm{C}$$. The temperature of the surrounding is $$10^{\circ} \mathrm{C}$$. The temperature of the body after the next 7 minutes will be:
$$34^{\circ} \mathrm{C}$$
$$28^{\circ} \mathrm{C}$$
$$32^{\circ} \mathrm{C}$$
$$30^{\circ} \mathrm{C}$$

Explanation

Newton's law of cooling states that the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings. The average rate of cooling can be represented as:

$$\frac{T_1-T_2}{t} = K \left(\frac{T_1+T_2}{2} - T_s\right)$$

where:

  • $T_1$ and $T_2$ are the initial and final temperatures of the body,
  • $t$ is the time it takes for the body to cool from $T_1$ to $T_2$,
  • $T_s$ is the temperature of the surroundings, and
  • $K$ is a constant of proportionality.

In the first 7 minutes, the body cools from $60^\circ C$ to $40^\circ C$, and the surrounding temperature is $10^\circ C$. So, the first equation is:

$$\frac{60-40}{7} = K \left(\frac{60+40}{2} - 10\right) \tag{1}$$

In the next 7 minutes, the body cools from $40^\circ C$ to $T^\circ C$, with the same surrounding temperature. So, the second equation is:

$$\frac{40-T}{7} = K \left(\frac{40+T}{2} - 10\right) \tag{2}$$

Solving equation (1) for $K$ and substituting into equation (2) will give the temperature $T$ after the next 7 minutes:

$$T = 28^\circ C$$

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