Density of water at 4 °C and 20 °C are $$1000 kg/m^{3}\text{ and }998kg/m^{3}$$ respectively. The increase in internal energy of 4 kg of water when it is heated from 4 °C to 20 °C is_____ J.
(specific heat capacity of water = $$4.2\times\ 10^3$$J / kg K. and 1 atmospheric pressure $$=10^{5}Pa$$)

We need to find the increase in internal energy of 4 kg of water heated from 4 °C to 20 °C.

Given that the mass of water is $$m = 4$$ kg, the density at 4 °C is $$\rho_1 = 1000$$ kg/m³, the density at 20 °C is $$\rho_2 = 998$$ kg/m³, the specific heat capacity is $$c = 4.2 \times 10^3$$ J/(kg·°C), the temperature change is $$\Delta T = 20 - 4 = 16\ ^\circ\mathrm{C}$$, and the atmospheric pressure is $$P = 10^5$$ Pa.

Using the first law of thermodynamics, the change in internal energy is given by $$\Delta U = Q - W$$.

Since the heat supplied is $$Q = mc\Delta T = 4 \times 4200 \times 16 = 268800\text{ J},$$ we have $$Q = 268800\text{ J}.$$

At constant atmospheric pressure, the work done by the system is $$W = P\,\Delta V = P\Bigl(\frac{m}{\rho_2} - \frac{m}{\rho_1}\Bigr).$$

Substituting the values gives

$$W = 10^5\Bigl(\frac{4}{998} - \frac{4}{1000}\Bigr)

= 10^5\Bigl(\frac{4\times1000 - 4\times998}{998\times1000}\Bigr)

= 10^5 \times \frac{8}{998000}

= \frac{8 \times 10^5}{998000}

\approx 0.8\text{ J}.$$

Therefore, the internal energy change is

$$\Delta U = Q - W = 268800 - 0.8 = 268799.2\text{ J}.$$

Thus, the increase in internal energy is Option 3: 268799.2 J.