To find the change in internal energy, we need to consider both the heat added during the process and the work done during the process.

First, let's calculate the heat added ($Q$) to convert 1 kg of water at 100°C into steam at 100°C using the latent heat of vaporization:

$$Q = m \times L$$

where $$m = 1 \mathrm{~kg}$$ and $$L = 2257 \mathrm{~kJ/kg}$$:

$$Q = 1 \mathrm{~kg} \times 2257 \mathrm{~kJ/kg} = 2257 \mathrm{~kJ}$$

Next, let's calculate the work done ($W$) on the system during the process. The work done is given by:

$$W = -P \Delta V$$

where $$P$$ is the atmospheric pressure and $$\Delta V$$ is the change in volume. We are given that the atmospheric pressure is $$P = 1 \times 10^5 \mathrm{~Pa}$$, and the change in volume is

$$\Delta V = 1.671 \mathrm{~m}^3 - 1.00 \times 10^{-3} \mathrm{~m}^3 = 1.670 \mathrm{~m}^3$$.

Now, we can calculate the work done:

$$W = -(1 \times 10^5 \mathrm{~Pa})(1.670 \mathrm{~m}^3) = -167000 \mathrm{~J} = -167 \mathrm{~kJ}$$

Finally, we can find the change in internal energy ($\Delta U$) using the first law of thermodynamics:

$$\Delta U = Q + W$$

Substitute the values of $$Q$$ and $$W$$:

$$\Delta U = 2257 \mathrm{~kJ} - 167 \mathrm{~kJ} = 2090 \mathrm{~kJ}$$

The change in internal energy of the system during the process is +2090 kJ.