A .20-kg ice cube at 0.0°C has sufficient thermal energy added to it through the process of heating to cause total melting,and the resulting water is heated to $70.0 ^ { \circ } \mathrm { C }$ How much energy 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.

Question

Quizplus · Accepted Answer

The correct answer is A because the total energy added is the sum of the energy required to melt the ice and the energy required to heat the resulting water from 0.0°C to 70.0°C. The energy required to melt the ice is given by:Q = mL_F = (0.20 kg)(334,000 J/kg) = 66,800 JThe energy required to heat the resulting water from 0.0°C to 70.0°C is given by:Q = mcΔT = (0.20 kg)(4186 J/kg °C)(70.0°C) = 59,564 JTherefore, the total energy added is:Q = Q_melt + Q_heat = 66,800 J + 59,564 J = 126,364 J ≈ 130 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 $70.0 ^ { \circ } \mathrm { C }$