Heat And Thermodynamics Ques 57
- Two identical beakers $A$ and $B$ contain equal volumes of two different liquids at $60^{\circ} \mathrm{C}$ each and left to cool down. Liquid in $A$ has density of $8 \times 10^2 \mathrm{~kg} / \mathrm{m}^3$ and specific heat of $2000 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$ while liquid in $B$ has density of $10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ and specific heat of $4000 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$. Which of the following best describes their temperature versus time graph schematically? (Assume the emissivity of both the beakers to be the same)
(2019 Main, 8 April I)
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Answer:
Correct Answer: 57.(b)
Solution:
- Key Idea
From Newton’s law of cooling, we have rate of cooling,
$$ \frac{d Q}{d t}=\frac{h}{m s}\left(T-T _0\right) $$
where, $h=$ heat transfer coefficient,
$T=$ temperature of body,
$T _0=$ temperature of surrounding,
$m=$ mass and $s=$ specific heat.
We know, $m=V \cdot \rho$
where, $V=$ volume and $\rho=$ density.
So, we have
$$ \frac{d Q}{d t}=\frac{h}{m s}\left(T-T _0\right)=\frac{h\left(T-T _0\right)}{V \cdot \rho s} $$
Since, $h,\left(T-T _0\right)$ and $V$ are constant for both beaker.
$$ \therefore \quad \frac{d Q}{d t} \propto \frac{1}{\rho s} $$
We have given that $\rho _A=8 \times 10^{2} kgm^{-3}$,
$$ \begin{aligned} & \rho _B=10^{3} kgm^{-3}, \\ & s _A=2000 J kg^{-1} K^{-1} \text { and } \\ & s _B=4000 J kg^{-1} K^{-1}, \end{aligned} $$
$$ \begin{gathered} \rho _A s _A=16 \times 10^{5} \\ \text { and } \\ \text { So, } \rho _A<\rho _B, s _A=4 \times 10^{6} \\ \Rightarrow \quad \frac{1}{\rho _A s _A}>\frac{1}{\rho _B s _B} \Rightarrow \frac{d}{d t}>\frac{d Q _B}{d t} \end{gathered} $$
So, for container $B$, rate of cooling is smaller than the container $A$. Hence, graph of $B$ lies above the graph of $A$ and it is not a straight line (slope of $A$ is greater than $B$ ).
BETA
