Heat and Thermodynamics 3 Question 2
2. Two identical beakers $A$ and $B$ contain equal volumes of two different liquids at $60^{\circ} C$ each and left to cool down. Liquid in $A$ has density of $8 \times 10^{2} kg / m^{3}$ and specific heat of $2000 Jkg^{-1} K^{-1}$ while liquid in $B$ has density of $10^{3} kg m^{-3}$ and specific heat of $4000 J kg^{-1} K^{-1}$. Which of the following best describes their temperature versus time graph $A$ and $C$ are maintained at a temperature of $95^{\circ} C$ while the columns $B$ and $D$ are maintained at $5^{\circ} C$. The height of the liquid in $A$ and $D$ measured from the base line are $52.8 cm$ and $51 cm$ respectively. Determine the linear coefficient of thermal expansion of the liquid.
(1997, 5M)
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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 $B$ ).
