 When $1g$ of water at $100{}^\circ C$ gets converted into steam at the same temperature, the change in volume is approximately:A.$1$$ccB.1000$$cc$C.$1500$$ccD.1670$$cc$ Verified
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Hint: Before talking about the answer, we should know what steam is. Steam is defined as a water which is in gas form. It is formed due to evaporation. Here, we are going to find the change in the volume with the help of density formula.
Formula used: $m=\delta V$
where $m$ is the mass of water, $\delta$ is the density of water and $V$ is the volume of water.

When water is heated, it will start evaporating, that means it starts getting converted into water vapour and expands. At $100{}^\circ C$ , the water gets evaporated and converted into steam, which is an invisible gas.
Here, it is given that $1g$ of water at $100{}^\circ C$ gets converted into steam at the same temperature.
And as we know, density of water, $\delta =1000kg/{{m}^{3}}$
$\Rightarrow 1g/c{{m}^{3}}$
And density of steam, $\delta =0.6kg/{{m}^{3}}$
$\Rightarrow 0.0006g/c{{m}^{3}}$
Here, mass of water $(m)$ is $1g$
$m=\delta V$
Substituting the values we have, in the above formula, we get
$V=\dfrac{m}{\delta }$
$V=\dfrac{1}{1}=1$
$V$ denotes as the volume of water
Volume of water is $1cc$ .
Volume of the steam,
${V}'=\dfrac{m}{\rho }$
$V'=\dfrac{1}{0.0006}=1666.7cc\approx 1667cc$
$V'$ denotes as the volume of steam

Here, the density of steam is used.
Therefore, the change in volume,
$\Delta V={V}'-V \\ \Delta V=1667-1=1666cc \\$
which is close to $1670$$cc So, the correct option is (D), that is, 1670$$cc$ .

Make sure to convert $kg/{{m}^{3}}$ to $g/c{{m}^{3}}$. The volume of water and steam is calculated in $cm$ .
The mass of one $cc$ of water at $3.98{}^\circ C$ is nearly equal to one gram.