1kg of water at 100 ° C is converted into steam at 100 ° C by...

The work to be done in the process is given by

$\begin{cases}dW & = & PdV \\ = & 1 \times {10}^{5}Pa \times ( 1 . 671 - 0 . 001 ) {m}^{3} \\ = & 1 . 670 \times {10}^{5}J\end{cases}$

The change in heat energy during the vaporisation process can be calculated as follows-

$\begin{cases}\Delta{Q}_{\text{supplied}} & = & 2257 \times 1 \times {10}^{3}J \\ = & 22 . 57 \times {10}^{5}J\end{cases}$

Hence, the change in internal energy in the process is given by

$\begin{cases}\DeltaU & = & \Delta{Q}_{supplied} - \DeltaW \\ = & ( 22 . 57 - 1 . 67 ) \times {10}^{5}J \\ = & 20 . 9 \times {10}^{5}J \\ = & 2090kJ\end{cases}$