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 $70.0 ^ { \circ } \mathrm { C }$ How much heat is added? For water L_F = 334,000 J/kg, L_V = 2.256 × 10⁶ J/kg, the c = 4.186 x 10³ J/kg ∙ C. A) 130 kJ B) 14,000 kJ C) 81 kJ D) 59 kJ

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Quizplus · Accepted Answer

The heat required to melt the ice is:$$Q_1 = m L_F = (0.20 	ext{ kg})(334,000 	ext{ J/kg}) = 66,800 	ext{ J}$$The heat required to raise the temperature of the water is:$$Q_2 = mc_p \Delta T = (0.20 	ext{ kg})(4186 	ext{ J/kg} \cdot 	ext{C})(70.0 	ext{C} - 0.0 	ext{C}) = 59,000 	ext{ J}$$The total heat added is:$$Q = Q_1 + Q_2 = 66,800 	ext{ J} + 59,000 	ext{ J} = 125,800 	ext{ J} \approx 130 	ext{ kJ}$$

A 20-Kg Ice Cube at 0 70.0∘C70.0 ^ { \circ } \mathrm { C }70.0∘C

A 20-Kg Ice Cube at 0 $70.0 ^ { \circ } \mathrm { C }$