How much energy (heat)is required to convert 52.0 g of ice at -10.0°C to steam at 100°C? Specific heat of ice 2.09 J/g • °C
Specific heat of water 4.18 J/g • °C
Specific heat of steam 1.84 J/g • °C
Molar heat of fusion 6.02 kJ/mol
Molar heat of vaporization 40.7 kJ/mol

Question

Quizplus · Accepted Answer

To find the total energy required, calculate the energy for each step: heating ice, melting ice, heating water, vaporizing water, and heating steam. Add them together:  1. Heating ice from -10°C to 0°C: q = m * c * ΔT = 52.0 g * 2.09 J/g°C * 10°C = 1086.8 J  2. Melting ice: q = n * ΔH_fusion = (52.0 g / 18.015 g/mol) * 6.02 kJ/mol = 17.38 kJ  3. Heating water from 0°C to 100°C: q = m * c * ΔT = 52.0 g * 4.18 J/g°C * 100°C = 21736 J  4. Vaporizing water: q = n * ΔH_vaporization = (52.0 g / 18.015 g/mol) * 40.7 kJ/mol = 117.7 kJ  5. Heating steam from 100°C to 100°C: q = m * c * ΔT = 0 J (no temperature change)  Total energy = 1.0868 kJ + 17.38 kJ + 21.736 kJ + 117.7 kJ = 157.9 kJ ≈ 158 kJ.

How Much Energy (Heat)is Required to Convert 52