A 2294-kg sample of water at 0° C is cooled to $- 36 ^ { \circ } \mathrm { C } \text {, }$ and freezes in the process. How much heat is liberated? For water L_F = 334,000 J/kg and L_V= 2.256 × 10⁶ J/kg. The specific heat of ice is 2050 J/kg ? K. A)935,000 kJ B)597,000 kJ C)1,110,000 kJ D)334,000 kJ

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The heat liberated is the sum of the heat removed to cool the water to its freezing point, the heat of fusion, and the heat removed to cool the ice to -36°C.$$Q = mc_p (0 - 273) + mL_F + mc_i (-36 - 0)$$$$Q = (2294 kg)(4186 J/kg K)(-273 K) + (2294 kg)(334,000 J/kg) + (2294 kg)(2050 J/kg K)(-36 K)$$$$Q = -2.47 	imes 10^6 J + 766 	imes 10^6 J - 1.67 	imes 10^6 J$$$$Q = 935,000 kJ$$

A 2294-Kg Sample of Water at 0° C Is Cooled −36∘C, - 36 ^ { \circ } \mathrm { C } \text {, }−36∘C,

A 2294-Kg Sample of Water at 0° C Is Cooled $- 36 ^ { \circ } \mathrm { C } \text {, }$