Final answer:
When 1 gram of steam at 100°C condenses and cools to 22°C, a total of 2.58601 kJ of heat is released. This includes heat from both the condensation of steam to liquid water and the subsequent cooling of the water.
Step-by-step explanation:
When considering the amount of heat released during the condensation of steam, we utilize the concept of the heat of vaporization of water. For each mole of steam that condenses, 40.7 kJ of heat are released. Assuming the molar mass of water is approximately 18.015 g/mol, we can determine the moles in 1 gram of steam and calculate the heat released upon condensation and subsequent cooling to 22°C.
First, convert 1 gram of steam to moles: 1 gram ÷ 18.015 g/mol = 0.0555 moles. Then, calculate the heat released during condensation: 0.0555 moles × 40.7 kJ/mol = 2.25985 kJ.
After condensation, the water at 100°C must cool to 22°C. The specific heat capacity of water is approximately 4.18 J/g°C. The heat released during cooling can be calculated using, q = mcΔT, where m is the mass of the water, c is the specific heat capacity, and ΔT is the temperature change. This gives us: q = (1 g) × (4.18 J/g°C) × (100°C - 22°C) = 326.16 J or 0.32616 kJ. Finally, add the heat released during condensation to the heat released during cooling to obtain the total heat released.
The total heat released is 2.25985 kJ (condensation) + 0.32616 kJ (cooling) = 2.58601 kJ.