Heat is added to a 2.0 kg piece of ice at a rate of 793.0 kW.How long will it take for the ice at 0.0ºC to melt? (For water L_f = 334 kJ/kg and L_v = 2257 kJ/kg. )

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Quizplus · Accepted Answer

The heat required to melt the ice is:$$Q = mL_f = (2.0 	ext{ kg})(334 	ext{ kJ/kg}) = 668 	ext{ kJ}$$The rate at which heat is added is:$$P = 793.0 	ext{ kW} = 793.0 	ext{ kJ/s}$$Therefore, the time required to melt the ice is:$$t = \frac{Q}{P} = \frac{668 	ext{ kJ}}{793.0 	ext{ kJ/s}} = 0.84 	ext{ s}$$

Heat Is Added to a 2