A .20-kg ice cube at $0.0 ^ { \circ } \mathrm { C }$ 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 $L _ { F } = 334,000 \mathrm {~J} / \mathrm { kg } , L \mathrm {~V} = 2.256 \times 10 ^ { 6 }$ $\mathrm { J } / \mathrm { kq } _ { \text {, the } c } = 4.186 \times 10 ^ { 3 } \mathrm {~J} / \mathrm { kq } \cdot \mathrm { C } \text {. }$

Question

Quizplus · Accepted Answer

The total heat added is the sum of the heat required to melt the ice and the heat required to raise the temperature of the resulting water to $70.0 ^ { \circ } \mathrm { C }$. The heat to melt the ice is $Q_{melt} = m \cdot L_{F} = 0.20 \, \text{kg} \cdot 334,000 \, \text{J/kg} = 66,800 \, \text{J}$. The heat to raise the temperature of the water is $Q_{heat} = m \cdot c \cdot \Delta T = 0.20 \, \text{kg} \cdot 4,186 \, \text{J/kg} \cdot \mathrm { C } \cdot (70.0 - 0.0) ^ { \circ } \mathrm { C } = 58,604 \, \text{J}$. Adding these together gives $66,800 \, \text{J} + 58,604 \, \text{J} = 125,404 \, \text{J}$, which is approximately 130 kJ.

A 20-Kg Ice Cube At At 0.0 0.0∘C0.0 ^ { \circ } \mathrm { C }0.0∘C

A 20-Kg Ice Cube At At 0.0 $0.0 ^ { \circ } \mathrm { C }$