A bowl filled with very hot soup cools from $98^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in 2 minutes when the room temperature is $22^{\circ} \mathrm{C}$. How long it will take to cool from $75^{\circ} \mathrm{C}$ to $69^{\circ} \mathrm{C}$ ?

• A

  
  2 minutes
• Correct answer
  B

  
  1.4 minutes
• C

  
  0.5 minute
• D

  
  1 minute

Solution

The rate of heat flow is given by
$\begin{aligned}<br />& \frac{\Delta Q}{\Delta t}=-K\left(T-T_0\right) \Rightarrow \frac{\Delta Q}{\Delta t}=-K\left(T_{a v g}-T_0\right) \\<br />& \text { Initially } T_{a v g}=\frac{T_1+T_2}{2} \\<br />& \text { Here, } T_1=98^{\circ} \mathrm{C} \\<br />& T_2=86^{\circ} \mathrm{C}<br />\end{aligned}$
$\begin{aligned} & \text { (i) } \frac{\mathrm{ms} \times 12}{2}=-\mathrm{K}\left(\frac{98+86}{2}-22\right) \\ & 6=-\frac{\mathrm{K}}{\mathrm{ms}}\left[\frac{98+86}{2}-22\right] \quad(\because \Delta \mathrm{Q}=\mathrm{ms} \Delta \mathrm{T})\end{aligned}$


Now, cool from $\mathrm{T}_1^{\prime}=75^{\circ} \mathrm{C}$
$\begin{aligned}<br />& \mathrm{T}_2=69^{\circ} \mathrm{C} \\<br />& \mathrm{T}_{\mathrm{avg}}=\frac{\mathrm{T}_1^{\prime}+\mathrm{T}_2^{\prime}}{2}=\frac{75+69}{2}<br />\end{aligned}$
(ii) $\frac{\mathrm{ms} \times 6}{\Delta \mathrm{t}}=-\mathrm{K}\left(\frac{75+69}{2}-22\right)$

Divide equation (ii) by (i), we have
$\begin{aligned}<br />& \frac{6}{\Delta t(6)}=\frac{50}{70} \\<br />& \Delta t=\frac{7}{5}=1.4 \mathrm{~min}<br />\end{aligned}$

Hence, the correct answer is option: 'B'.