How much energy is needed to convert 60.6 grams of ice at 0.00°C to water at 55.4°C?
Specific heat of ice = 2.10 J/(g°C)
Specific heat of water = 4.18 J/(g°C)
Heat of fusion = 333 J/g
Heat of vaporization = 2258 J/g
A) 56.4 kJ
B) 14.0 kJ
C) 151 kJ
D) 34.2 kJ
E) 6.13 kJ

Question

Quizplus · Accepted Answer

To find the total energy needed, we calculate the energy required for each step of the process: melting the ice to water at 0°C, and then heating the water from 0°C to 55.4°C.1. Melting ice to water (at 0°C):$$Q_{fusion} = m \times \Delta H_{fusion} = 60.6 \, \text{g} \times 333 \, \text{J/g} = 20179.8 \, \text{J}$$2. Heating water from 0°C to 55.4°C:$$Q_{water} = m \times c_{water} \times \Delta T = 60.6 \, \text{g} \times 4.18 \, \text{J/(g°C)} \times 55.4 \, \text{°C} = 13913.5 \, \text{J}$$Total energy required:$$Q_{total} = Q_{fusion} + Q_{water} = 20179.8 \, \text{J} + 13913.5 \, \text{J} = 34093.3 \, \text{J} \approx 34.1 \, \text{kJ}$$The closest answer is 34.2 kJ.

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