How much heat must be added to a 8.0 -kg block of ice at $- 8 ^ { \circ } \mathrm { C }$ to change it to water at $14 ^ { \circ } \mathrm { C } ?$ The
specific heat of ice is $2050 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { C } ^ { \circ }$ the specific heat of water is $4186 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { C } ^ { \circ }$the latent heat of fusion of ice is 334,000 J/kg, and 1cal=4.186 J.

Question

How much heat must be added to a  8.0 -kg block of ice at $- 8 ^ { \circ } \mathrm { C }$ to change it to water at $14 ^ { \circ } \mathrm { C } ?$ The
specific heat of ice is $2050 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { C } ^ { \circ }$ the specific heat of water is $4186 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { C } ^ { \circ }$the latent heat of  fusion of ice is  334,000 J/kg, and  1cal=4.186 J.

Quizplus · Accepted Answer

The total heat required is the sum of the heat to warm the ice to 0°C, the heat to melt the ice, and the heat to warm the water to 14°C. Calculating each:  1. Heat to warm ice: $Q_1 = mc\Delta T = 8 \times 2050 \times 8 = 131,200 \, \text{J}$.  2. Heat to melt ice: $Q_2 = mL_f = 8 \times 334,000 = 2,672,000 \, \text{J}$.  3. Heat to warm water: $Q_3 = mc\Delta T = 8 \times 4186 \times 14 = 469,024 \, \text{J}$.  Total heat $Q = Q_1 + Q_2 + Q_3 = 3,272,224 \, \text{J}$.  Convert to kcal: $3,272,224 \, \text{J} \times \frac{1 \, \text{kcal}}{4186 \, \text{J}} \approx 780 \, \text{kcal}$.

How Much Heat Must Be Added to a 8 −8∘C- 8 ^ { \circ } \mathrm { C }−8∘C

How Much Heat Must Be Added to a 8 $- 8 ^ { \circ } \mathrm { C }$