A 2294 kg sample of water at 0°C is cooled to -36ºC,and freezes in the process.How much heat is liberated? (For water L_f = 334 kJ/kg and L_v = 2257 kJ/kg.The specific heat for ice is 2050 J/kg · 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°C.$$Q = mc_p(0-(-36)) + mL_f + mc_s(-36-0)$$$$Q = (2294 kg)(4187 J/kg · K)(36 K) + (2294 kg)(334 kJ/kg) + (2294 kg)(2050 J/kg · K)(36 K)$$$$Q = 3.34 × 10^6 J + 7.66 × 10^5 J + 1.67 × 10^6 J$$$$Q = 5.78 × 10^6 J = 5780 kJ$$

A 2294 Kg Sample of Water at 0°C Is Cooled