How much heat must be added to a 8.0-kg block of ice at -8°C to change it to water at $14 ^ { \circ } \mathrm { C } ?$ The specific heat of ice is 2050 J/kg ? C°, the specific heat of water is 4186 J/kg ? C°, the latent heat of fusion of ice is 334,000 J/kg, and 1 cal = 4.186 J.
A)780 kcal
B)140 kcal
C)180 kcal
D)810 kcal
E)730 kcal

Question

Quizplus · Accepted Answer

Explanation:The heat required to raise the temperature of the ice from -8°C to 0°C is:$$Q_1 = mc_p \Delta T = (8.0 	ext{ kg})(2050 	ext{ J/kg } \degree 	ext{C})(8 \degree 	ext{C}) = 131,200 	ext{ J}$$The heat required to melt the ice at 0°C is:$$Q_2 = mL = (8.0 	ext{ kg})(334,000 	ext{ J/kg}) = 2,672,000 	ext{ J}$$The heat required to raise the temperature of the water from 0°C to 14°C is:$$Q_3 = mc_p \Delta T = (8.0 	ext{ kg})(4186 	ext{ J/kg } \degree 	ext{C})(14 \degree 	ext{C}) = 456,264 	ext{ J}$$The total heat required is:$$Q = Q_1 + Q_2 + Q_3 = 131,200 	ext{ J} + 2,672,000 	ext{ J} + 456,264 	ext{ J} = 3,259,464 	ext{ J}$$Converting to kcal:$$Q = 3,259,464 	ext{ J} 	imes \frac{1 	ext{ kcal}}{4186 	ext{ J}} = 780 	ext{ kcal}$$

How Much Heat Must Be Added to a 8 14∘C?14 ^ { \circ } \mathrm { C } ?14∘C?

How Much Heat Must Be Added to a 8 $14 ^ { \circ } \mathrm { C } ?$