Determine the increase in the internal energy of 10 g of water as...

Mass of water, $m = 10 \text{ g} = 0.01 \text{ kg}$

Heat required to raise the temperature of water from 0^{}^{{}^{\circ}}\text{C} to 100^{}^{{}^{\circ}}\text{C} is given by:
$Q_1 = m c \Delta T = 0.01 \times 4200 \times 100 = 4200 \text{ J}$

Heat required to convert water at 100^{}^{{}^{\circ}}\text{C} into steam is given by:
$Q_2 = mL = 0.01 \times 2.25 \times 10^6 = 22500 \text{ J}$

Total heat supplied to the system is:
$Q = Q_1 + Q_2 = 4200 + 22500 = 26700 \text{ J}$

Work done by the system against the constant external pressure during vaporization is:
$W = P(V_{\text{steam}} - V_{\text{water}})$

Neglecting the initial volume of water, the volume of steam is:
$V_{\text{steam}} = \dfrac{m}{\rho_{\text{steam}}} = \dfrac{0.01}{0.6} = \dfrac{1}{60} \text{ m}^3$

Substituting the values to find the work done:
$W = 100 \times 10^3 \times \dfrac{1}{60} = 1666.67 \text{ J}$

Using the first law of thermodynamics, the increase in internal energy is:
$\Delta U = Q - W$
$\Delta U = 26700 - 1666.67 = 25033.33 \text{ J}$

$\Delta U \approx 2.50 \times 10^4 \text{ J}$