Q138 / 160 · TS EAMCET 2018 (04 May Shift 2) · Physics
A container is filled with a liquid that cools from $100{ }^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in $5 \mathrm{~min}$, when kept at room temperature of $30^{\circ} \mathrm{C}$. The time that it must have taken to cool down to $80^{\circ} \mathrm{C}$ from its initial temperature approximately is
- A$1.7 \mathrm{~min}$
- Correct answerB$2.6 \mathrm{~min}$
- C$8.2 \mathrm{~min}$
- D$4.1 \mathrm{~min}$
Solution
Given that, In first case,
$$
\begin{array}{c|c}
\mathrm{T}_1=100^{\circ} \mathrm{C} & \mathrm{T}_0=30^{\circ} \mathrm{C} \\
\mathrm{T}_2=70^{\circ} \mathrm{C} & \text { time, } t_1=5 \mathrm{~min}=300 \mathrm{~s}
\end{array}
$$
According to Newton's law of cooling,
$$
\begin{aligned}
m C\left(\frac{T_1-T_2}{t}\right) & =k\left(\frac{T_1+T_2}{2}-T_0\right) \\
m C\left[\frac{100-70}{300}\right] & =k\left[\frac{100+70}{2}-30\right] \\
m C\left[\frac{30}{300}\right] & =k\left[\frac{170}{2}-30\right] \\
m C\left[\frac{1}{10}\right] & =k[85-30] \\
m C\left(\frac{1}{10}\right) & =k(55)
\end{aligned}
$$
For, second case,
$$
\begin{array}{c|c}
T_1=100^{\circ} \mathrm{C} & \mathrm{T}_0=30^{\circ} \mathrm{C} \\
\mathrm{T}_2=80^{\circ} \mathrm{C} & \text { time, } t_2=t^{\prime} \min =t^{\prime \prime} \mathrm{s}
\end{array}
$$
According to Newton's law of cooling,
$$
\begin{aligned}
m C\left[\frac{100-80}{t^{\prime \prime}}\right] & =k\left[\frac{100+80}{2}-30\right] \\
m C\left[\frac{20}{t^{\prime \prime}}\right] & =k[90-30] \\
\Rightarrow \quad m C\left[\frac{20}{t^{\prime \prime}}\right] & =k[60]
\end{aligned}
$$
After, solving Eqs. (i) and (ii), we get $t^{\prime \prime}=3.05 \mathrm{~min}$ (nearest answer is 2)
Hence, the correct answer is option: 'B'.