QJEEM23Phc531: JEE Main 2023 Physics - Class 11 JEEM23Phc531 Solution: A body cools in 7 minutes from to . The temperature of the…

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:

$T = 28^\circ C$