The latent heat of vaporisation of water is $2240 \mathrm{~J}$. If the work done in the process of vaporisation of $1 \mathrm{~g}$ is $168 \mathrm{~J}$, the increase in internal energy is

According to the first law of thermodynamics, the heat supplied $Q$ is equal to the sum of the change in internal energy $\Delta U$ and the work done $W$ by the system during the process of vaporisation.

$Q = \Delta U + W$

Given that the latent heat of vaporisation $L = 2240 \text{ J}$ for $1 \text{ g}$ of water, the heat supplied $Q$ is $2240 \text{ J}$.

The work done $W$ during the process is given as $168 \text{ J}$.

Substituting these values into the equation:

$2240 \text{ J} = \Delta U + 168 \text{ J}$

Solving for the change in internal energy $\Delta U$:

$\Delta U = 2240 \text{ J} - 168 \text{ J}$

$\Delta U = 2072 \text{ J}$

Answer: 2072 J