Q163 / 180 · COMEDK 2014 · Physics
A pan filled with hot food cools from $94^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in $2 \mathrm{~min}$, when the room temperature is $20^{\circ} \mathrm{C}$. The time taken for the food to cool from $86^{\circ} \mathrm{C}$ to $74^{\circ} \mathrm{C}$ will be
- A$500 \mathrm{~s}$
- B$420 \mathrm{~s}$
- C$200 \mathrm{~s}$
- Correct answerD$210 \mathrm{~s}$
Solution
According to Newton's law of cooling,
$$
\frac{T_{1}-T_{2}}{t}=K\left(\frac{T_{1}+T_{2}}{2}-T_{s}\right) ...(i)
$$
where, $T_{s}$ is the temperature of surrounding.
For the first case, $T_{1}=94^{\circ} \mathrm{C}, T_{2}=86^{\circ} \mathrm{C}$
$$
t=2 \min =120 \mathrm{~s}, T_{s}=20^{\circ} \mathrm{C}
$$
$\therefore$ Putting these values in Eq. (i), we get
$\Rightarrow$
$$
\begin{aligned}
\frac{94-86}{120} &=K\left(\frac{94+86}{2}-20\right) \\
K &=\frac{8}{120 \times 70} ...(ii)
\end{aligned}
$$
For the second case, $T_{1}=86^{\circ} \mathrm{C}, T_{2}=74^{\circ} \mathrm{C}$
$\therefore$ From Eq. (i), we get
$$
\begin{aligned}
\frac{86-74}{t} =K\left(\frac{86+74}{2}-20\right) \\
\Rightarrow \quad \frac{12}{t} =\frac{8}{120 \times 70}(60) \\
\Rightarrow \quad t =210 \mathrm{~s}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.