How much energy must be added to a 8.0-kg block of ice at -8°C to change it to water at $14 ^ { \circ } \mathrm { C } \text { ? }$ 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.

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Explanation:The energy required to change the ice to water at 0°C is:$$Q_1 = m c_i (0 - (-8)) = (8.0 	ext{ kg})(2050 	ext{ J/kg} \cdot 	ext{C})(8 	ext{C}) = 131,200 	ext{ J}$$The energy required to melt the ice at 0°C is:$$Q_2 = mL_f = (8.0 	ext{ kg})(334,000 	ext{ J/kg}) = 2,672,000 	ext{ J}$$The energy required to raise the temperature of the water from 0°C to 14°C is:$$Q_3 = m c_w (14 - 0) = (8.0 	ext{ kg})(4186 	ext{ J/kg} \cdot 	ext{C})(14 	ext{C}) = 459,264 	ext{ J}$$The total energy required is:$$Q = Q_1 + Q_2 + Q_3 = 131,200 	ext{ J} + 2,672,000 	ext{ J} + 459,264 	ext{ J} = 3,262,464 	ext{ J}$$Converting to kcal:$$Q = 3,262,464 	ext{ J} 	imes \frac{1 	ext{ kcal}}{4186 	ext{ J}} = 780 	ext{ kcal}$$

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

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