I) Specific heat capacity of a body is the amount of heat energy required to raise the temperature of 1 kg of the substance by 1 Kelvin (or 1 degree Celsius).

Given:
Mass of copper ball, $m_c = 50 \text{ g} = 0.050 \text{ kg}$
Initial temperature of copper ball (from boiling water), $T_c = 100^\circ\text{C}$
Specific heat capacity of copper, $c_c = 400 \text{ Jkg}^{-1}\text{K}^{-1}$
Initial temperature of water, $T_w = 23^\circ\text{C}$
Final temperature of mixture, $T_f = 62^\circ\text{C}$
Specific heat capacity of water, $c_w = 4200 \text{ Jkg}^{-1}\text{K}^{-1}$

According to the principle of calorimetry, heat lost by the copper ball equals the heat gained by the water.

$$Q_{\text{lost by copper}} = Q_{\text{gained by water}}$$

The formula for heat exchange is $Q = mc\Delta T$.

The change in temperature for copper is $\Delta T_c = T_c - T_f$.

$$\Delta T_c = 100^\circ\text{C} - 62^\circ\text{C} = 38^\circ\text{C}$$ $$Q_{\text{lost by copper}} = m_c c_c \Delta T_c$$ $$Q_{\text{lost by copper}} = (0.050 \text{ kg}) \times (400 \text{ Jkg}^{-1}\text{K}^{-1}) \times (38 \text{ K})$$ $$Q_{\text{lost by copper}} = 760 \text{ J}$$

The change in temperature for water is $\Delta T_w = T_f - T_w$.

$$\Delta T_w = 62^\circ\text{C} - 23^\circ\text{C} = 39^\circ\text{C}$$ $$Q_{\text{gained by water}} = m_w c_w \Delta T_w$$ $$Q_{\text{gained by water}} = m_w \times (4200 \text{ Jkg}^{-1}\text{K}^{-1}) \times (39 \text{ K})$$ $$Q_{\text{gained by water}} = m_w \times 163800 \text{ Jkg}^{-1}$$

$$760 \text{ J} = m_w \times 163800 \text{ Jkg}^{-1}$$ $$m_w = \frac{760 \text{ J}}{163800 \text{ Jkg}^{-1}}$$ $$m_w \approx 0.0046398 \text{ kg}$$ $$m_w \approx 4.64 \text{ g}$$

The mass of water in the container is $\boxed{\text{4.64 g}}$.