Calculate the amount of heat energy required to convert 55.0 g of water at 62.5^oC to steam at 124.0 $\degree$C.(C_water = 4.18 J/g $\degree$C; C_steam = 2.02 J/g $\degree$C; molar heat of vaporization of liquid water = 4.07 x 10⁴ J/mol)

Question

Quizplus · Accepted Answer

To find the total heat energy required, calculate the energy for each step: heating water to 100°C, vaporizing it, and heating steam to 124°C. First, use q = mcΔT for heating water: q1 = 55.0 g * 4.18 J/g°C * (100°C - 62.5°C). Then, use the heat of vaporization: q2 = (55.0 g / 18.015 g/mol) * 4.07 x 10^4 J/mol. Finally, use q = mcΔT for heating steam: q3 = 55.0 g * 2.02 J/g°C * (124°C - 100°C). Add q1, q2, and q3, and convert to kJ. The total is approximately 1.35 x 10^2 kJ.

Calculate the Amount of Heat Energy Required to Convert 55 °\degree°

Calculate the Amount of Heat Energy Required to Convert 55 $\degree$