A body at a temperature of ${50}^{\circ }\mathrm{C}$ cools to ${49}^{\circ }\mathrm{C}$ in time Δt when it is placed in a room maintained at $-3^{\circ }\mathrm{C}$. The same body cools from ${50}^{\circ }\mathrm{C}$ to ${49}^{\circ }\mathrm{C}$ in time Δt ' when room temperature is ${24}^{\circ }\mathrm{C}$. Assume heat loss through radiation only and the specific heat capacity of the body remains constant with change in temperature. $\mathrm{\Delta }{t}^{\mathrm{\text{'}}}=P\times \mathrm{\Delta }t$
Then nearest integer to 2 x P is $\left( \right.$Take $(270/297{)}^3=3/4$)

$\Rightarrow \mathrm{ }-\frac{dT}{dt}=\left(\frac{4e\sigma A{T}_0^3}{ms}\right)\mathrm{\Delta }T\mathrm{ }\Rightarrow \mathrm{ }-\frac{dT}{dt}=k\mathrm{\Delta }T$

This is Newton's law of cooling. We need to take care that the constant k depends on $T_0^3$. When room temperature is $-3^{\circ }\mathrm{C}$ and ${24}^{\circ }\mathrm{C}$ let the value of constant be $k_1$ and $k_2$.

Then $\frac{k_2}{k_1}={\left(\frac{297}{270}\right)}^3=(1.1{)}^3=1.33$

$\therefore$When ball is in first room

$\frac{1^{0}C}{\mathrm{\Delta }t}=k_1[49.5-(-3\left)\right]$

[49.5 = average temperature of the body during cooling]
In second room $\frac{1^{0}C}{\mathrm{\Delta }{t}^{\mathrm{\text{'}}}}=k_2[49.5-24]$

$$\begin{gathered} \therefore \mathrm{ }\left(i\right)÷\left(ii\right)\mathrm{ }\frac{\mathrm{\Delta }{t}^{\mathrm{\text{'}}}}{\mathrm{\Delta }t}=\frac{k_1}{k_2}\times \frac{52.5}{25.5} \\ \therefore \mathrm{ }\mathrm{\Delta }{t}^{\mathrm{\text{'}}}=\frac{1}{1.33}\times \frac{52.5}{25.5}=\mathrm{\Delta }t=1.55\mathrm{\Delta }t \end{gathered}$$