For water, the latent heat of freezing is 334 kJ/kg and the specific heat capacity averages 4.19 kJ/kgK. The quantity of heat to be removed from 1 kg of water at 30°C in order to turn it into ice at 0°C is:

Correct option is D

​$$\begin{aligned}&\text{Use the formula:} \\&Q_1 = m \cdot c \cdot \Delta T \\[1em]&\text{Where:} \\&m = 1\, \text{kg} \\&c = 4.19\, \text{kJ/kg} \cdot \text{K} \\&\Delta T = 30 - 0 = 30\, \text{K} \\[1em]&Q_1 = 1 \cdot 4.19 \cdot 30 = 125.7\, \text{kJ} \\[2em]&\textbf{Freezing water at 0}^\circ\text{C to ice at 0}^\circ\text{C} \\[0.5em]&\text{Use the latent heat formula:} \\&Q_2 = m \cdot L_f \\[1em]&\text{Where:} \\&L_f = 334\, \text{kJ/kg} \\[1em]&Q_2 = 1 \cdot 334 = 334\, \text{kJ} \\[2em]&\textbf{Total Heat to be Removed:} \\&Q_{\text{total}} = Q_1 + Q_2 = 125.7 + 334 = \boxed{459.7\, \text{kJ}}\end{aligned}$$​​