1 kg of water at 100 ° C is converted into steam at 100 ° C by boiling at atmospheric pressure. The volume of water changes from 1 . 00 × 10 - 3 m 3 as a liquid to 1 . 671 m 3 as steam. The change in

Question

1 kg of water at 100 ° C is converted into steam at 100 ° C by boiling at atmospheric pressure. The volume of water changes from 1 . 00 × 10 - 3 m 3 as a liquid to 1 . 671 m 3 as steam. The change in internal energy of the system during the process will be (Given latent heat of vaporisation = 2257 kJ / kg , Atmospheric pressure = 1 × 10 5 Pa

TestHub · Accepted Answer

The work to be done in the process is given by

$\begin{array}{rcl} d W & = & P d V \\ = & 1 \times 10^{5} Pa \times (1.671 - 0.001) m^{3} \\ = & 1.670 \times 10^{5} J \end{array}$

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

$\begin{array}{rcl} Δ Q_{supplied} & = & 2257 \times 1 \times 10^{3} J \\ = & 22.57 \times 10^{5} J \end{array}$

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

$\begin{array}{rcl} Δ U & = & Δ Q_{supplied} - Δ W \\ = & (22.57 - 1.67) \times 10^{5} J \\ = & 20.9 \times 10^{5} J \\ = & 2090 kJ \end{array}$

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