Q. The density of water at $20^{\circ} C$ is $998\, kg\, m ^{-3}$ and that at $40^{\circ} C$ is $992\, kg\, m ^{-3}$. The co-efficient of cubical expansion of water is :

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Solution:

We use $\rho^{\prime} =\rho(1-\gamma \Delta T)$
$998 =\rho(1-20 \gamma)$ ...(1)
$992 =\rho(1-40 \gamma)$ ...(2)
Dividing (1) by (2), we get
$\frac{499}{496} =\frac{1-20 \gamma}{1-40 \gamma}$
$499-19960 \gamma =496-9920 \gamma$
$10040 \gamma =3$
$\gamma =2.988 \times 10^{-4}\,{}^{\circ} C ^{-1}$