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 \(30^{\circ} \mathrm{C}\) ]

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):
\(\frac{10}{(\Delta t) 5}=\frac{20}{40}\)
\(\Rightarrow \Delta t=4 \mathrm{~min}\)