Final answer:
When 5.4 g of steam at 105°C condenses and cools to 100°C, approximately 12.21 kJ of heat is released. The heat released during the condensation contributes to the total heat change.
Step-by-step explanation:
To calculate the heat released when 5.4 g of steam at 105°C is condensed to liquid water at 100°C, we must consider two processes: the condensation of steam to water and the cooling of steam from 105°C to 100°C.
Step 1: Cooling of steam
Heat loss for cooling steam from 105°C to 100°C is negligible because the temperature change is minimal (the specific heat of steam is not provided in the question).
Step 2: Condensation of steam to water
Based on the provided data, the molar mass of water (H₂O) is approximately 18.015 g/mol, and the latent heat of vaporization is 40.7 kJ/mol. Given the mass of steam as 5.4 g, we first convert the mass to moles:
Moles of steam = Mass of steam / Molar mass of water = 5.4 g / 18.015 g/mol = 0.3 mol
The heat released during the condensation of steam to water at 100°C can be calculated by multiplying the moles of steam by the molar heat of vaporization:
Heat released = Moles of steam × Molar heat of vaporizationHeat released = 0.3 mol × 40.7 kJ/mol = 12.21 kJ
Therefore, 12.21 kJ of heat is released during the process, and since the question specifies a negative number for the release of heat, the answer is -12.21 kJ.