JEE Physics 2023 Question with Solution

Given: The body cools from $60^\circ \text{C}$ to $40^\circ \text{C}$ in $6$ minutes, and the surrounding temperature is $10^\circ \text{C}$.

Find: The temperature after the next $6$ minutes.

Using Newton’s law of cooling, the average rate of cooling is taken as proportional to the excess temperature over surroundings.

For the first interval,

$$\frac{60-40}{6} = k\left[\frac{60+40}{2} - 10\right]$$

So,

$$k = \frac{20}{6 \times 40} = \frac{1}{12}$$

For the second interval, let the temperature after the next $6$ minutes be $T$. Then,

$$\frac{40-T}{6} = k\left[\frac{40+T}{2} - 10\right]$$

Substituting $k = \frac{1}{12}$,

$$\frac{40-T}{6} = \frac{1}{12}\left[\frac{40+T}{2} - 10\right]$$

Solving this gives

$$T = 28^\circ \text{C}$$

Therefore, the temperature after the next $6$ minutes is $28^\circ \text{C}$.

Given: Two equal time intervals of $6$ minutes each are considered.

Find: The final temperature in the second interval.

From the solution working:

$$\frac{60-40}{6} = k\left[\frac{60+40}{2} - 10\right]$$

The average temperature during the first interval is $\frac{60+40}{2} = 50$, so excess over surroundings is $50-10 = 40$. Hence,

$$\frac{20}{6} = 40k$$ $$k = \frac{20}{240} = \frac{1}{12}$$

Now for the second interval,

$$\frac{40-T}{6} = k\left[\frac{40+T}{2} - 10\right]$$

Substitute $k = \frac{1}{12}$:

$$\frac{40-T}{6} = \frac{1}{12}\left(\frac{40+T}{2} - 10\right)$$

Multiply by $12$:

$$2(40-T) = \frac{40+T}{2} - 10$$ $$80 - 2T = \frac{40+T}{2} - 10$$

Multiply by $2$:

$$160 - 4T = 40 + T - 20$$ $$160 - 4T = 20 + T$$ $$140 = 5T$$ $$T = 28$$

Therefore, the required temperature is $28^\circ \text{C}$.