A bucket full of hot water is kept in a room. If it cools from $75^{\circ} \mathrm{C}$ to $70^{\circ}…

A bucket full of hot water is kept in a room. If it cools from $75^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in $\mathrm{t}_1$ minutes, from $70^{\circ} \mathrm{C}$ to $65^{\circ} \mathrm{C}$ in $\mathrm{t}_2$ minutes and $65^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in $\mathrm{t}_3$ minutes, then
  1. $\mathrm{t}_1 \lt \mathrm{t}_2 \lt \mathrm{t}_3$
  2. $t_1 \gt t_2 \gt t_3$
  3. $t_1=t_2=t_3$
  4. $\mathrm{t}_1 \lt \mathrm{t}_2=\mathrm{t}_3$

Solution

According to Newton's law of cooling, Rate of cooling $\propto$ Mean temperature difference $\Rightarrow \frac{\text { Fall in temperature }}{\text { Time }(\mathrm{t})} \propto\left(\frac{\theta_1+\theta_2}{2}-\theta_0\right)$ Case 1: $\left(\frac{\theta_1+\theta_2}{2}\right)_1=\left(\frac{75+70}{2}\right)_1=72.5$ Case 2: $\left(\frac{\theta_1+\theta_2}{2}\right)_2=\left(\frac{70+65}{2}\right)_2=67.5$ Case 3: $\left(\frac{\theta_1+\theta_2}{2}\right)_3=\left(\frac{65+60}{2}\right)_3=62.5$ $\begin{aligned} \therefore \quad & \left(\frac{\theta_1+\theta_2}{2}\right)_1\gt\left(\frac{\theta_1+\theta_2}{2}\right)_2\gt\left(\frac{\theta_1+\theta_2}{2}\right)_3 \\ & \Rightarrow \mathrm{t}_1 \lt \mathrm{t}_2 \lt \mathrm{t}_3 \end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 1)

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