A .20 kg ice cube at 0.0°C has sufficient heat added to it to cause total melting,and the resulting water is heated to 70.0ºC.How much heat is added? (For water L_f = 334 kJ/kg and L_v= 2257 kJ/kg. ) A)130 kJ B)14,000 kJ C)81 kJ D)59 kJ

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Explanation:Heat required to melt ice = m L_f = (0.20 kg)(334 kJ/kg) = 66.8 kJHeat required to raise the temperature of water = m c_p ΔT = (0.20 kg)(4.19 kJ/kg°C)(70.0°C) = 59.5 kJTotal heat added = 66.8 kJ + 59.5 kJ = 126.3 kJ ≈ 130 kJ

A 20 Kg Ice Cube at 0