Exam

BITSAT

Year

2023

Paper

Paper 2

Subject

Physics

Chapter

Thermal Properties of Matter

BITSAT 2023 · Paper 2 · Physics · Thermal Properties of Matter

A bowl filled with very hot soup cools from 98^ ^ C to 86^ ^ C in 2...

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 $69^{{}^{\circ}} \mathrm{C}$ ?

Step-by-step solution

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


Now, cool from $\mathrm{T}_1^{\prime}=75^{\circ}$$\mathrm{C}$$<br />$$\begin{aligned} \\ & \mathrm{T}_2=69^{\circ} \mathrm{C} \\ \\ & \mathrm{T}_{\mathrm{avg}}=\frac{\mathrm{T}_1^{\prime}+\mathrm{T}_2^{\prime}}{2}=\frac{75+69}{2} \\ \end{aligned}$$<br />(ii)$\frac{\mathrm{ms} \times 6}{\Delta $$\mathrm{t}$$}=-$$\mathrm{K}$$ \left($\frac{75+69}{2}-$22\right)$<img src="/api/proxy-image?url=https$$\begin{aligned}\\ & \frac{6}{\Delta t(6)}=\frac{50}{70}\\ \\ & \Delta t=\frac{7}{5}=1.4 \mathrm{~min} \\ \end{aligned}$$