Question

When 52gm of a liquid in a vessel is heated from ${0^\circ }C$ to ${100^\circ }C$ , $2gm$ of liquid is expelled. If $104gm$ of liquid is taken in a vessel made of same material, the mass of the liquid expelled on heating from ${0^\circ }C$ to${100^\circ }C$ :
A) $4gm$
B) $2gm$
C) $8gm$
D) $1gm$

Answer 1

Hint: We can solve this question in two steps first be take when 52gm liquid get heated then due to expansion in liquid $2gm$ of liquid expelled from here we calculate coefficient of thermal expansion of liquid with respect to container
And in second step we take when $104gm$ liquid get heated
Coefficient of thermal expansion remains the same in both because the container is made of the same material and the liquid is also the same.

Complete step by step solution:
When heat given to liquid then liquid as well as container both will expand how much the liquid expand with respect to container that much the liquid expelled out from container
Mass expelled of liquid = expansion in liquid with respect to container
$ \Rightarrow \Delta m = m{\gamma _{lc}}\left( {\Delta T} \right)$
Here${\gamma _{lc}}$ is the coefficient of thermal expansion of liquid with respect to container
$\Delta m$ Is the mass expelled of liquid
$\Delta T$ Is the temperature difference
So we can find the coefficient of thermal expansion of liquid with respect to container
$ \Rightarrow {\gamma _{lc}} = \dfrac{{\Delta m}}{{m\left( {\Delta T} \right)}}$
Putting the value $\Delta m = 2gm$ $m = 52gm$ and $\Delta T = {\left( {100 - 0} \right)^\circ }C$
$ \Rightarrow {\gamma _{lc}} = \dfrac{2}{{52\left( {100 - 0} \right)}}$
$ \Rightarrow {\gamma _{lc}} = \dfrac{2}{{52 \times 100}}$ ............. (1)
Step 2
Now in the second time we want to find a mass of liquid expelled when initially taking $104gm$ liquid.
From above formula coefficient of thermal expansion of liquid with respect to container
$ \Rightarrow {\gamma _{lc}} = \dfrac{{\Delta m}}{{m\left( {\Delta T} \right)}}$
Put given value $m = 104gm$ and $\Delta T = {\left( {100 - 0} \right)^\circ }C$
$ \Rightarrow {\gamma _{lc}} = \dfrac{{\Delta m}}{{104\left( {100} \right)}}$ ................... (2)
Because we neither change the liquid nor material of container so coefficient of thermal expansion of liquid with respect to container is same for both cases
So (1) = (2)
$ \Rightarrow \dfrac{2}{{52\left( {100} \right)}} = \dfrac{{\Delta m}}{{104\left( {100} \right)}}$
Solving this we can find $\Delta m$ mass of liquid expelled
$ \Rightarrow \Delta m = \dfrac{{2 \times 104}}{{52}}$
$ \Rightarrow \Delta m = \dfrac{{208}}{{52}}$
$\therefore \Delta m = 4gm$
Hence in the second case mass of the liquid expelled is $8gm$.

Hence option (C) is correct.

Note: Coefficient of thermal expansion of liquid with respect to container can be explain as the difference between coefficient of thermal expansion of liquid and coefficient of thermal expansion of container
$ \Rightarrow {\gamma _{lc}} = {\gamma _l} - {\gamma _c}$
Where ${\gamma _l}$ is the coefficient of thermal expansion of container material,
${\gamma _l}$ Is the coefficient of thermal expansion of liquid.