Question

The density of water at $20^\circ C$ is $998kg/{m^3}$ and at $40^\circ C$ is $992kg/{m^3}$. The coefficient of volume expansion of water is
(A) ${10^{ - 4}}/^\circ C$
(B) $3 \times {10^{ - 4}}/^\circ C$
(C) $2 \times {10^{ - 4}}/^\circ C$
(D) $6 \times {10^{ - 4}}/^\circ C$

Answer 1

Density of a given substance is inversely proportional to the volume of the substance. The volume at lower temperature is used to calculate the coefficient of volume expansion.

Formula used: In this solution we will be using the following formulae;
$\Rightarrow$ $\gamma = \dfrac{{\Delta V}}{{{V_i}\Delta T}}$ where $\gamma$ is the coefficient of volume expansion, $\Delta V$ is the change in volume from initial temperature to final temperature, and ${V_i}$ is the initial volume (or volume at the lower temperature), and $\Delta T$ is the change in temperature.
$\Rightarrow$ $V = \dfrac{m}{\rho }$ where $V$ is volume, $m$ is mass and $\rho$ is density.

Complete Step-by-Step Solution:
To calculate the coefficient of volume expansion of water, we recall the formula
$\Rightarrow$ $\gamma = \dfrac{{\Delta V}}{{{V_i}\Delta T}}$ where $\gamma$ is the coefficient of volume expansion, $\Delta V$ is the change in volume from initial temperature to final temperature, and ${V_i}$ is the initial volume (or volume at the lower temperature), and $\Delta T$ is the change in temperature.
But $V = \dfrac{m}{\rho }$ where $m$ is mass and $\rho$ is density.
Hence,
$\Rightarrow$ $\dfrac{{\Delta V}}{V} = - \dfrac{{\Delta \rho }}{\rho }$
Hence,
$\Rightarrow$ $\gamma = \dfrac{{ - \Delta \rho }}{{{\rho _i}\Delta T}}$
Hence, by inserting known values we have
$\Rightarrow$ $\gamma = \dfrac{{ - \left( {992 - 998} \right)}}{{992\left( {40 - 20} \right)}}$
Hence, by computation,
$\Rightarrow$ $\gamma = \dfrac{6}{{992\left( {20} \right)}} = 3 \times {10^{ - 4}}/^\circ C$

Hence, the correct option is B

Note: For clarity, the relationship $\dfrac{{\Delta V}}{V} = - \dfrac{{\Delta \rho }}{\rho }$ can be gotten from $V = \dfrac{m}{\rho }$ as follows,
We first differentiate the volume with respect to the density, we have
$\Rightarrow$ $\dfrac{{dV}}{{d\rho }} = - \dfrac{m}{{{\rho ^2}}}$
Then by multiplying both sides by $d\rho$ and rewriting the equation, we have
$\Rightarrow$ $dV = - \dfrac{m}{\rho }\dfrac{{d\rho }}{\rho }$
Now, since $\dfrac{m}{\rho } = V$, then by substituting into the equation above, we have
$dV = - V\dfrac{{d\rho }}{\rho }$
Then,
$\Rightarrow$ $\dfrac{{dV}}{V} = - \dfrac{{d\rho }}{\rho }$
Hence, $\dfrac{{\Delta V}}{V} = - \dfrac{{\Delta \rho }}{\rho }$
Also, note that in the formula for coefficient of expansion, the volume at the lower temperature (initial volume) is used. Mistakes are often made when it concerns cooling, that students often use the initial volume as the volume at the hotter temperature (because it’s initial in terms of cooling). Nonetheless, because the coefficient of expansion is being measured, it must be taken as though it was being heated.