Heat And Thermodynamics Ques 3
- A sinker of weight $w _ 0$ has an apparent weight $w _ 1$ when placed in a liquid at a temperature $T _ 1$ and $w _ 2$ when weighed in the same liquid at a temperature $T _ 2$. The coefficient of cubical expansion of the material of the sinker is $\beta$. What is the coefficient of volume expansion of the liquid?
(1978)
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Answer:
Correct Answer: 3.$\beta _ L=\beta\left(\frac{w _ 0-w _ 1}{w _ 0-w _ 2}\right)+\frac{\left(w _ 2-w _ 1\right)}{\left(w _ 0-w _ 2\right)\left(T _ 2-T _ 1\right)}$
Solution:
Formula:
Calorimetry And Thermal Expansion Types Of Thermometers :
- Apparent weight $=$ actual weight - upthrust $ w _ 1=w _ 0-F _ 1 $
Here, $F _ 1=$ upthrust at temperature $T _ 1$
$=\left(\right.$ volume of sinker at temperature $\left.T _ 1\right) \times$ (density of liquid at temperature $T _ 1$ ) $\times g$
$ =\frac{w _ 0}{g} \times \frac{1}{\left(\rho _ s\right) _ {T _ 1}} \times g \times\left(\rho _ l\right) _ {T _ 1}=w _ 0\left[\frac{\left(\rho _ l\right) _ {T _ 1}}{\left(\rho _ s\right) _ {T _ 1}}\right] $
$\therefore \quad w _ 1=w _ 0-w _ 0\left[\frac{\left(\rho _ l\right) _ {T _ 1}}{\left(\rho _ s\right) _ {T _ 1}}\right]$ Similarly, $w _ 2=w _ 0-w _ 0\left[\frac{\left(\rho _ l\right) _ {T _ 2}}{\left(\rho _ s\right) _ {T _ 2}}\right]$
From Eqs. (i) and (ii), we have
$ \frac{\left(\rho _ l\right) _ {T _ 1}}{\left(\rho _ s\right) _ {T _ 2}}=\frac{w _ 0-w _ 1}{w _ 0} \text { and } \frac{\left(\rho _ l\right) _ {T _ 2}}{\left(\rho _ s\right) _ {T _ 2}}=\frac{w _ 0-w _ 2}{w _ 0} $
From these two equations we have,
$ \frac{\left(\rho _ I\right) _ {T _ 1}}{\left(\rho _ I\right) _ {T _ 2}} \times \frac{\left(\rho _ s\right) _ {T _ 2}}{\left(\rho _ s\right) _ {T _ 1}}=\frac{w _ 0-w _ 1}{w _ 0-w _ 2} $
Using, $\quad \frac{\rho^{\prime}}{\rho}=\frac{1}{1+\beta \Delta T}$
Here, $\beta=$ thermal coefficient of volume expansion
$ \therefore \quad \frac{1+\beta _ I\left(T _ 2-T _ 1\right)}{1+\beta _ s\left(T _ 2-T _ 1\right)}=\frac{w _ 0-w _ 1}{w _ 0-w _ 2} $
Given, $\beta _ s=\beta$
$ \therefore \quad\left(\frac{w _ 0-w _ 2}{w _ 0-w _ 1}\right)\left[1+\beta _ l\left(T _ 2-T _ 1\right)\right]=1+\beta\left(T _ 2-T _ 1\right) $
Solving this equation, we get
$\beta _ I =\frac{ {1+\beta (T _ 2-T _ 1 ) } {\frac{w _ 0-w _ 1}{w _ 0-w _ 2} }-1}{ (T _ 2-T _ 1 )} $
$ =\beta\left(\frac{w _ 0-w _ 1}{w _ 0-w _ 2}\right)+\frac{w _ 2-w _ 1}{\left(w _ 0-w _ 2\right)\left(T _ 2-T _ 1\right)}$
BETA
