A 2294-kg sample of water at 0° C is cooled to $- 36 ^ { \circ } \mathrm { C }$ 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

Question

Quizplus · Accepted Answer

The heat liberated includes the heat released during freezing and the heat released when the temperature of ice is lowered. First, calculate the heat released during freezing: Q1 = m * L_F = 2294 kg * 334,000 J/kg = 766,196,000 J. Then, calculate the heat released when the temperature of ice is lowered: Q2 = m * c * ΔT = 2294 kg * 2050 J/kg∙K * 36 K = 169,083,600 J. Total heat liberated: Q = Q1 + Q2 = 935,279,600 J = 935,280 kJ ≈ 935,000 kJ.

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

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