Calculate the heat that is either gainer or lost when 25.0 g of steam ( MW=18.02, cp: .0753, delta H fus 6.01 kj, delta H vap:40.67 kj, at 100C condenses to liquid water at 100 C.

To calculate the heat gained or lost when steam condenses to liquid water, we need to consider several factors:

1. Heat required to raise the temperature of steam from 0°C to 100°C: This can be calculated using the formula Q = m × Cp × ΔT, where Q is the heat energy, m is the mass of the substance, Cp is the specific heat capacity, and ΔT is the change in temperature. For steam, the specific heat capacity (Cp) is 0.0753 J/g°C.

Q1 = m × Cp × ΔT
= 25.0 g × 0.0753 J/g°C × (100 - 0)°C
= 188.25 J

So, 188.25 J of heat energy is required to raise the temperature of 25.0 grams of steam from 0°C to 100°C.

2. Heat required to condense the steam to liquid water at 100°C: To convert steam to liquid water at the boiling point, we need to consider the heat of vaporization (ΔHvap). For steam, the heat of vaporization is 40.67 kJ/g.

Q2 = m × ΔHvap
= 25.0 g × 40.67 kJ/g
= 1016.75 kJ
= 1016.75 × 1000 J

So, 1016.75 kJ (or 1016.75 × 1000 J) of heat energy is required to condense 25.0 grams of steam to liquid water at 100°C.

3. Total heat transferred: Since the question asks for the total heat gained or lost when steam condenses to liquid water, we need to sum up the heat from the previous steps.

Total heat transferred = Q1 + Q2
= 188.25 J + 1016.75 × 1000 J
= 188.25 J + 1,016,750 J
= 1,016,938.25 J

Therefore, the total heat gained or lost when 25.0 grams of steam condenses to liquid water at 100°C is approximately 1,016,938.25 J.