An object is cooled from $75^{\circ} \mathrm{C}$ to $65^{\circ} \mathrm{Cin} 2 \mathrm{~min}$. The time, it…
An object is cooled from $75^{\circ} \mathrm{C}$ to $65^{\circ} \mathrm{Cin} 2 \mathrm{~min}$. The time, it takes to cool from $55^{\circ} \mathrm{C}$ to $45^{\circ} \mathrm{Cis}$ [The temperature of surrounding is $30^{\circ} \mathrm{C}$ ]
$9 \mathrm{~min}$
$10 \mathrm{~min}$
$4 \mathrm{~min}$
$8 \mathrm{~min}$
Solution
Taking the newton's law of cooling in average form:
$\frac{\Delta Q}{\Delta t}=-K\left(T_{a v g}-T_0\right)$, where $T_{a v}=\frac{T_1+T_2}{2}, K$ is the thermal conductivity.
Using, $\Delta Q=m C_P \Delta T$ the heat and heat capacity relation:
$\Rightarrow \frac{m C_P \Delta T}{\Delta t}=-K\left(T_{a v g}-T_0\right)$
Case (1) Given soup cools from $75^{\circ} \mathrm{C}$ to $65^{\circ} \mathrm{C}$ in 2 minutes when the room temperature is $30^{\circ} \mathrm{C}$ :
$\Rightarrow \frac{m C_P \times 10^{\circ} \mathrm{C}}{2 \min }=-K\left\{\frac{(75+65)^{\circ} \mathrm{C}}{2}-30^{\circ} \mathrm{C}\right\}$
Case (2) Let the soup cools from cool from $55^{\circ} \mathrm{C}$ to $45^{\circ} \mathrm{C}$ int minutes when the room temperature is $30^{\circ} \mathrm{C}$ :
$\Rightarrow \frac{m C_P \times 10^{\circ} \mathrm{C}}{\Delta t}=-K\left\{\frac{(55+45)^{\circ} \mathrm{C}}{2}-30^{\circ} \mathrm{C}\right\}$
On taking the ratio of equation (2) \& (1):
$\frac{10}{(\Delta t) 5}=\frac{20}{40}$
$\Rightarrow \Delta t=4 \mathrm{~min}$
Case (2) Let the soup cools from cool from $55^{\circ} \mathrm{C}$ to $45^{\circ} \mathrm{C}$ int minutes when the room temperature is $30^{\circ} \mathrm{C}$ :
$\Rightarrow \frac{m C_P \times 10^{\circ} \mathrm{C}}{\Delta t}=-K\left\{\frac{(55+45)^{\circ} \mathrm{C}}{2}-30^{\circ} \mathrm{C}\right\}$
On taking the ratio of equation (2) \& (1):
$\frac{10}{(\Delta t) 5}=\frac{20}{40}$
$\Rightarrow \Delta t=4 \mathrm{~min}$