Question

$ 45gm $ of alcohol is needed to completely fill up a weight thermometer at $ 15^\circ C $ . Find the weight of alcohol which will overflow when the weight thermometer is heated to $ 33^\circ C $ . ( $ {\gamma _a} = 121 \times {10^{ - 5}}^\circ {C^{ - 1}} $ )
(A) $ 0.96gm $
(B) $ 0.9gm $
(C) $ 1gm $
(D) $ 2gm $

Answer 1

Hint
With the rise in temperature, alcohol filled in the thermometer expands and overflows out of the container. This amount of overflow can tell the difference in temperature. Reverse is also possible when change in the temperature is given.
 $ \Delta V = {V_0}\left( {{\gamma _L} - {\gamma _C}} \right)\Delta T $
Where $ \Delta V $ is the change in volume that takes place due to thermal expansion.
 $ {V_0} $ is the initial volume of the liquid.
 $ {\gamma _L} $ is defined as the coefficient of volumetric expansion of the liquid (here alcohol)
 $ {\gamma _C} $ is the coefficient of volumetric expansion for the container (not specified here)
 $ \Delta T $ is the change in temperature, and can be calculated by subtracting initial temperature from the final temperature.

Complete step by step answer
The amount of alcohol which overflows is used to determine its temperature raise. Thus, the increase in volume must be equal to the volume of alcohol which falls on the weighing scale.
Therefore,
 $ \Delta V = {V_0}\left( {{\gamma _L} - {\gamma _C}} \right)\Delta T $
We know that, $ \dfrac{m}{\rho } = V $
Assuming that the change in density is very small, we can rewrite the equation as-
 $ \dfrac{{\Delta m}}{\rho } = \dfrac{{{m_0}\left( {{\gamma _L} - {\gamma _C}} \right)\Delta T}}{\rho } $
 $ \Delta m = {m_0}\left( {{\gamma _L} - {\gamma _C}} \right)\Delta T $
Also, assuming that the container of the alcohol does not expand with rise in temperature, the equation becomes-
 $ \Delta m = {m_0}{\gamma _L}\Delta T $
Where $ \Delta m $ is the mass of alcohol which reaches the weighing scale.
 $ {m_0} $ is the total mass of the alcohol before the expansion, which is $ 45g $
It is given, $ {\gamma _L} = {\gamma _a} = 121 \times {10^{ - 5}} $
 $ \Delta T = 33 - 15 = 18^\circ C $
By putting the values we get-
 $ \Delta m = 45 \times 121 \times {10^{ - 5}} \times 18 $
 $ \Delta m = 98010 \times {10^{ - 5}} $
 $ \Delta m = 0.98010 \simeq 1 $
The mass of overflowing alcohol is $ 1gm $ . Thus option (C) is correct.

Note
 In practical aspects of this device, the density does not remain constant with respect to the temperature, In fact it is the feature which causes the expansion of liquid. An accurate formula to determine the overflowing weight is-
 $ \Delta m = \dfrac{{{V_0}({\gamma _L} - {\gamma _C})\Delta T{\rho _{L0}}}}{{1 + {\gamma _C}\Delta T}} $
Here, the term $ {\rho _{L0}} $ is the initial density of liquid, $ {\gamma _L} $ and $ {\gamma _C} $ are the coefficient of volumetric expansion for the liquid and the container, $ \Delta T $ is the temperature difference and $ {V_0} $ is the initial volume.