Determine the change in the internal energy of 2 kg of water when...

The heat supplied to the water is given by $\Delta Q = m c \Delta T$.

Substituting the given values:

$\Delta Q = 2 \times 4200 \times (4 - 0) = 33600 \text{ J}$

The change in volume of the water is \Delta V = V_2 - V_1 = m \left(\dfrac{1}{\rho_2} - \dfrac{1}{\rho_1} \right).

Substituting the densities:

\Delta V = 2 \left(\dfrac{1}{1000} - \dfrac{1}{999.9} \right) =2 \left(\dfrac{999.9 - 1000}{1000 \times 999.9} \right) = \dfrac{-0.2}{999900} \approx -$2 \times 10^{-7} \text{ m}^3$

The work done by the water against the atmospheric pressure is $\Delta W = P \Delta V$.

$\Delta W = 10^5 \times (-2 \times 10^{-7}) = -0.02 \text{ J}$

According to the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$.

$\Delta U = \Delta Q - \Delta W = 33600 - (-0.02) = 33600.02 \text{ J}$