question_answer

\[1.0{{m}^{3}}\]of water is converted into\[1671{{m}^{3}}\]of steam at atmospheric pressure and \[100{}^\circ C\]temperature. The latent heat of vapourisation of water is\[2.3\times {{10}^{6}}J/kg\]. If 2.0 kg of water be converted into steam at atmospheric pressure and \[100{}^\circ C\] temperature, then how much will be the increase in its internal energy (in joule)? Density of water is \[1.0\text{ }\times {{10}^{3}}\text{ }kg/{{m}^{3}}\], atmospheric pressure\[=1.01\times {{10}^{5}}N/{{m}^{2}}.\]

Answer:

\[\left( 4.263\text{ }\times \text{ }{{10}^{3}} \right)\]