A 2294-kg sample of water at $0 ^ { \circ } \mathrm { C }$ is cooled to $- 36 ^ { \circ } \mathrm { C }$ and freezes in the process. How much heat is liberated? For water $L F = 334,000 \mathrm {~J} / \mathrm { kg } \text { and } L V = 2.256 \times 10^6 \mathrm {~J} / \mathrm { kg }$ The specific heat of ice is 2050 $\mathrm { J } / \mathrm { kg } \cdot \mathrm { K }$

Question

Quizplus · Accepted Answer

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 ^ { \circ } \mathrm { C }$$$Q = mc_p (0 - 20) + mL_f + mc_i (-36 - 0)$$$$Q = (2294 \mathrm { kg } )(4187 \mathrm { J } / \mathrm { kg } \cdot \mathrm { K } )(-20 \mathrm { K }) + (2294 \mathrm { kg } )(334,000 \mathrm { J } / \mathrm { kg }) + (2294 \mathrm { kg } )(2050 \mathrm { J } / \mathrm { kg } \cdot \mathrm { K })(-36 \mathrm { K })$$$$Q = -1.88 \times 10^6 \mathrm { J } + 767 \times 10^6 \mathrm { J } - 1.68 \times 10^6 \mathrm { J }$$$$Q = 935,000 \mathrm { kJ }$$

A 2294-Kg Sample of Water At 0∘C0 ^ { \circ } \mathrm { C }0∘C

A 2294-Kg Sample of Water At $0 ^ { \circ } \mathrm { C }$