Question 63: Properties of Matter

Question

A body cools from $$80^{\circ} \mathrm{C}$$ to $$60^{\circ} \mathrm{C}$$ in 5 minutes. The temperature of the surrounding is $$20^{\circ} \mathrm{C}$$. The time it takes to cool from $$60^{\circ} \mathrm{C}$$ to $$40^{\circ} \mathrm{C}$$ is :

Accepted Answer

We are given the rate of cooling is proportional to the temperature difference between the body and the surroundings. Mathematically, it can be expressed as:
$$\frac{\Delta T_{body}}{\Delta t} \propto (T_{body} - T_{surroundings})$$
Here,
$$\Delta T_{body}$$ is the change in temperature of the body,
$$\Delta t$$ is the time taken for the change in temperature,
$$T_{body}$$ is the temperature of the body,
$$T_{surroundings}$$ is the temperature of the surroundings.
Now, introducing a proportionality constant $$c$$, we can write:
$$\frac{\Delta _{body}}{\Delta t} = c (T_{body} - T_{surroundings})$$
For the first cooling interval (from $$80^{\circ} \mathrm{C}$$ to $$60^{\circ} \mathrm{C}$$ in 5 minutes):
$$\frac{20}{5} = c (70 - 20)$$
For the second cooling interval (from $$60^{\circ} \mathrm{C}$$ to $$40^{\circ} \mathrm{C}$$ in $$x$$ minutes):
$$\frac{20}{x} = c (50 - 20)$$
Now, we have two equations:
1) $$\frac{20}{5} = c (50)$$
2) $$\frac{20}{x} = c (30)$$
Solve equation (1) for $$c$$:
$$c = \frac{20}{5 \cdot 50} = \frac{1}{25}$$
Substitute the value of $$c$$ into equation (2):
$$\frac{20}{x} = \frac{1}{25}(30)$$
Solve for $$x$$:
$$x = \frac{20 \cdot 25}{30} = \frac{500}{30} = \frac{25}{3}$$
So, the time it takes for the body to cool from $$60^{\circ} \mathrm{C}$$ to $$40^{\circ} \mathrm{C}$$ is $$\frac{25}{3}$$ minutes, which is equal to 500 seconds.

Answer

500 s

Answer

$$\frac{25}{3} \mathrm{~S}$$

Answer

450 s

Answer

420 s

Properties of Matter

A body cools from $80^{\circ} \mathrm{C} to 60^{\circ} \mathrm{C} in 5 minutes. The temperature of the surrounding is 20^{\circ} \mathrm{C}. The time it takes to cool from 60^{\circ} \mathrm{C} to 40^{\circ} \mathrm{C}$ is :

Choose the correct answer: