1 kg of water at 100$$^\circ$$C is converted into steam at 100$$^\circ$$C by boiling at atmospheric pressure. The volume of water changes from $$1.00 \times 10^{-3}$$ 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 $$= 1 \times 10^5$$ Pa)

We use the first law of thermodynamics: $$\Delta U = Q - W$$.

The heat absorbed during vaporisation is:

$$Q = mL = 1 \times 2257 = 2257 \text{ kJ}$$

Now, the work done during expansion at constant pressure is:

$$W = P\Delta V = P(V_2 - V_1)$$

$$W = 1 \times 10^5 \times (1.671 - 1.00 \times 10^{-3})$$

$$W = 1 \times 10^5 \times 1.670 = 1.670 \times 10^5 \text{ J} = 167 \text{ kJ}$$

So the change in internal energy is:

$$\Delta U = Q - W = 2257 - 167 = 2090 \text{ kJ}$$

Hence, the answer is $$+2090$$ kJ.