TS EAMCET 5-Aug-2021 Shift 1 Question Paper

Given, In case 1 , body cools from temperature $70^{\circ}$ to $40^{\circ}\,\mathrm{C}$ i.e. $T_{1}=70^{\circ}\,\mathrm{C}, T_{2}=40^{\circ}\,\mathrm{C}$ and time $t_{1}=5\,\mathrm{min}$ In case 2 , temperature reduces from $60^{\circ}\,\mathrm{C}$ to $40^{\circ}\,\mathrm{C}$ in time $(t_{2})$ i.e, $T_{3}=60^{\circ}\,\mathrm{C}, T_{4}=40^{\circ}\,\mathrm{C}, t_{2}=$ ? Consider temperature of surrounding, $T_{s}=20^{\circ}\,\mathrm{C}$. By using Newton's law of cooling, $\frac{T_i-T_f}{t}=K\left(\frac{T_i+T_f}{2}-T_s\right)$. . . (i) where, $T_{i}, T_{f}$ be initial and final temperature of body, $t$ is the time taken by body to reach $T_{i}$ to $T_{f}$ and $K$ is thermal heat coefficient. ccording to first case, By using Eq. (i), we get $\frac{T_1-T_2}{t_1}=K\left(\frac{T_1+T_2}{2}-T_s\right)$ $\therefore \frac{70-40}{5}=K\left(\frac{70+40}{2}-20\right)$ $\Rightarrow \frac{30}{5}=K\left(\frac{110}{2}-20\right) \Rightarrow 6=K(35)$ $\Rightarrow K=\frac{6}{35}$ According to second case, again by using Eq. (i), we get $\frac{T_3-T_4}{t_2}=K\left(\frac{T_3+T_4}{2}-T_s\right)$ $\frac{60-40}{t_2}=\frac{6}{35}\left(\frac{60+40}{2}-20\right)$ $\Rightarrow \frac{60-40}{t_2}=\frac{6}{35}\times 30 \Rightarrow \frac{20}{t_2}=\frac{6}{7}\times 6$ $t_2=\frac{20\times 7}{36}=\frac{140}{36}=3.89\,\mathrm{min}$