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 :

<p>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:</p> <p>$\frac{\Delta T_{body}}{\Delta t} \propto (T_{body} - T_{surroundings})$</p> <p>Here, </p> <ul> <li>$\Delta T_{body}$ is the change in temperature of the body,</li> <li>$\Delta t$ is the time taken for the change in temperature,</li> <li>$T_{body}$ is the temperature of the body,</li> <li>$T_{surroundings}$ is the temperature of the surroundings.</li> </ul> <p>Now, introducing a proportionality constant $c$, we can write:</p> <p>$\frac{\Delta _{body}}{\Delta t} = c (T_{body} - T_{surroundings})$</p> <p>For the first cooling interval (from $80^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in 5 minutes):</p> <p>$\frac{20}{5} = c (70 - 20)$</p> <p>For the second cooling interval (from $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ in $x$ minutes):</p> <p>$\frac{20}{x} = c (50 - 20)$</p> <p>Now, we have two equations:</p> <p>1) $\frac{20}{5} = c (50)$ 2) $\frac{20}{x} = c (30)$</p> <p>Solve equation (1) for $c$:</p> <p>$c = \frac{20}{5 \cdot 50} = \frac{1}{25}$</p> <p>Substitute the value of $c$ into equation (2):</p> <p>$\frac{20}{x} = \frac{1}{25}(30)$</p> <p>Solve for $x$:</p> <p>$x = \frac{20 \cdot 25}{30} = \frac{500}{30} = \frac{25}{3}$</p> <p>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.</p>