SATHEE: Heat and Thermodynamics 3 Question 2

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 $8 \times 10^{2} k g / m^{3}$ and specific heat of $2000 J k g^{- 1} K^{- 1}$ while liquid in $B$ has density of $10^{3} k g m^{- 3}$ and specific heat of $4000 J k g^{- 1} K^{- 1}$ . 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 $52.8 c m$ and $51 c m$ 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,

$\frac{d Q}{d t} = \frac{h}{m s} (T - T_{0})$

where, $h =$ heat transfer coefficient,

$T =$ temperature of body,

$T_{0} =$ temperature of surrounding,

$m =$ mass and $s =$ specific heat.

We know, $m = V \cdot ρ$

where, $V =$ volume and $ρ =$ density.

So, we have

$\frac{d Q}{d t} = \frac{h}{m s} (T - T_{0}) = \frac{h (T - T_{0})}{V \cdot ρ s}$

Since, $h, (T - T_{0})$ and $V$ are constant for both beaker.

$∴ \frac{d Q}{d t} \propto \frac{1}{ρ s}$

We have given that $ρ_{A} = 8 \times 10^{2} k g m^{- 3}$ ,

$\begin{aligned} ρ_{B} = 10^{3} k g m^{- 3}, \\ s_{A} = 2000 J k g^{- 1} K^{- 1} and \\ s_{B} = 4000 J k g^{- 1} K^{- 1}, \end{aligned}$

$\begin{matrix} ρ_{A} s_{A} = 16 \times 10^{5} \\ and \\ So, ρ_{A} < ρ_{B}, s_{A} = 4 \times 10^{6} \\ \Rightarrow \frac{1}{ρ_{A} s_{A}} > \frac{1}{ρ_{B} s_{B}} \Rightarrow \frac{d}{d t} > \frac{d Q_{B}}{d t} \end{matrix}$

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 $B$ ).