When 500 kg of water is heated from 20⁰C to 100⁰C, then the increase in mass of the water will be [UP CPMT 2013)

Given, m = 500 kg

Rise in temperature,   $dT=\left({100}^{\circ }-{20}^{\circ }\right)\mathrm{C}={80}^{\circ }\mathrm{C}$

Specific heat of water,   $c={10}^3\mathrm{cal}{\mathrm{kg}}^{-1\circ }{\mathrm{C}}^{-1}$

Heat energy gained by water,   $E=\mathrm{mcd}T$

$=500\times {10}^3\times 80\mathrm{cal}=4\times {10}^7\times 42\mathrm{J}=16.8\times {10}^7\mathrm{J}$

From   $E=m{c}^2$

Increase in mass of water,

$m=\frac{E}{c^2}=\frac{16.8\times {10}^7}{{\left(3\times {10}^8\right)}^2}$

$=1.87\times {10}^{-9}\mathrm{kg}$