AP EAPCET 24-Aug-2021 Shift 1 Paper

Given that, for first condition, Initial temperature, $T_{1}=52.5^{\circ}\mathrm{C}$ Final temperature, $T_{v}=47.5^{\circ}\mathrm{C}$ Average temperature, $T_{\mathrm{av}}=\frac{T_{1}+T_{2}}{2}$ $=\frac{525+47.5}{2}=50^{\circ}\mathrm{C}$ For second condition, Initial temperature, $T_{1}=47.5^{\circ}\mathrm{C}$ Final temperature, $T_{2}=42.5^{\circ}\mathrm{C}$ Average temperature, $T_{\mathrm{av}}=\frac{T_{1}+T_{2}}{2}$ $=\frac{47.5+42.5}{2}=45^{\circ}\mathrm{C}$ Let $T_{0}$ be the temperature of surroundings. Now, using expression of rate of cooling, $R=\frac{\Delta T}{\Delta}=-K(T_{\mathrm{av}}-T_{0})$ Substituting the values in above equation from $1^{\text{st}}$ and 2nd conditions, we get $\frac{T_{1}-T_{2}}{t_{1}}=-K(T_{\mathrm{av}}-T_{0})$ $\frac{52.5-47.5}{5}=-K(50-T_{0})$ ...(i) Similarly, $\frac{47,5-42.5}{7.5}=-K(45-T_{0})$ ...(ii) Dividing Eq. (i) by Eq. (ii), we get $\frac{\frac{52.5-47.5}{5}}{\frac{47.5-42.5}{7.5}}=\frac{-K(50-T_{0})}{-K(45-T_{0})}$ $\Rightarrow \frac{5}{5} \times \frac{7.5}{5} = \frac{50-T_{0}}{45-T_{0}}$ $\Rightarrow \frac{3}{2} = \frac{50-T_{0}}{45-T_{0}}$ $T_{0}=35^{\circ}\mathrm{C}$ Hence, the temperature of surroundings is $35^{\circ}\mathrm{C}$.