JEE Physics 2023 Question with Solution

Given: The body cools from $80^{\circ}\text{C}$ to $60^{\circ}\text{C}$ in $300\,\text{s}$ and the surrounding temperature is $20^{\circ}\text{C}$.

Find: The time required to cool from $60^{\circ}\text{C}$ to $40^{\circ}\text{C}$.

Using Newton's law of cooling,

$$\frac{dT}{dt}=-k(T-T_s)$$

where $T$ is the temperature of the object, $T_s$ is the surrounding temperature, and $k$ is a positive constant.

For cooling from $80^{\circ}\text{C}$ to $60^{\circ}\text{C}$, the average temperature is

$$\frac{80+60}{2}=70^{\circ}\text{C}$$

Applying the relation,

$$\frac{60-80}{300}=-k(70-20)$$ $$\frac{-20}{300}=-50k$$ $$k=\frac{1}{750}$$

Now for cooling from $60^{\circ}\text{C}$ to $40^{\circ}\text{C}$, the average temperature is

$$\frac{60+40}{2}=50^{\circ}\text{C}$$

Applying the same relation,

$$\frac{40-60}{t_2}=-k(50-20)$$ $$\frac{-20}{t_2}=-30k$$

Substituting $k=\frac{1}{750}$,

$$t_2=\frac{20}{30\times \frac{1}{750}}=500\,\text{s}$$

Therefore, the time it takes to cool from $60^{\circ}\text{C}$ to $40^{\circ}\text{C}$ is $500\,\text{s}$. The correct option is A.

Given: First cooling interval: $80^{\circ}\text{C}$ to $60^{\circ}\text{C}$ in $5$ minutes. Surrounding temperature: $20^{\circ}\text{C}$.

Find: Second cooling interval time from $60^{\circ}\text{C}$ to $40^{\circ}\text{C}$.

From the extracted working, Newton's law of cooling is used in average-temperature form.

First interval:

$$T_1=80^{\circ}\text{C},\quad T_2=60^{\circ}\text{C},\quad t_1=300\,\text{s},\quad T_s=20^{\circ}\text{C}$$

Average temperature:

$$\frac{T_1+T_2}{2}=\frac{80+60}{2}=70^{\circ}\text{C}$$

Hence,

$$\frac{T_2-T_1}{t_1}=-k\left(\frac{T_1+T_2}{2}-T_s\right)$$ $$\frac{60-80}{300}=-k(70-20)$$ $$\frac{-20}{300}=-50k$$ $$k=\frac{20}{300\cdot 50}=\frac{1}{750}$$

Second interval:

$$T_1=60^{\circ}\text{C},\quad T_2=40^{\circ}\text{C},\quad T_s=20^{\circ}\text{C}$$

Average temperature:

$$\frac{60+40}{2}=50^{\circ}\text{C}$$

Now,

$$\frac{40-60}{t_2}=-k(50-20)$$ $$\frac{-20}{t_2}=-30k$$ $$t_2=\frac{20}{30k}$$

Substituting $k=\frac{1}{750}$,

$$t_2=\frac{20}{30\cdot \frac{1}{750}}=\frac{20\cdot 750}{30}=500\,\text{s}$$

Therefore, the required time is $500\,\text{s}$ and the correct option is A.