A 2,294-kg sample of water at 0° C is cooled to -36 ° 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.

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 = mass × L_F = 2,294 kg × 334,000 J/kg = 766,196,000 J. Then, calculate the heat released when the temperature of ice is lowered: Q2 = mass × specific heat of ice × temperature change = 2,294 kg × 2050 J/kg∙K × 36 K = 169,663,800 J. Total heat liberated = Q1 + Q2 = 935,859,800 J = 935,000 kJ.

A 2,294-Kg Sample of Water at 0° C Is Cooled