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 °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 cube is:Q1 = mLf = (0.20 kg)(334,000 J/kg) = 66,800 JThe heat required to raise the temperature of the water from 0.0°C to 70.0°C is:Q2 = mcΔT = (0.20 kg)(4186 J/kg°C)(70.0°C) = 58,604 JThe total heat added is:Q = Q1 + Q2 = 66,800 J + 58,604 J = 125,404 J ≈ 130 kJ

A 20-Kg Ice Cube at 0