How much heat in kJ is lost when 100.0g of steam at 200 deg C is cooled down until it forms ice at -30.0 deg C?

I'll refer you to your previous post.

http://www.jiskha.com/display.cgi?id=1331082239

To find the amount of heat lost when steam is cooled down to form ice, we need to determine the energy required for each phase change.

The first step is to calculate the heat lost during the cooling of steam from 200°C to 100°C. We can use the specific heat capacity formula:

Q = mcΔT

Where:
Q = heat lost or gained
m = mass
c = specific heat capacity
ΔT = change in temperature

The specific heat capacity of water vapor (steam) is approximately 2.03 J/(g°C). However, since the values given in the question are in kJ and grams, we will convert the specific heat capacity to kJ/(g°C).

Specific heat capacity of steam = 2.03 J/(g°C) = 0.00203 kJ/(g°C)

Now, we can calculate the heat lost during the cooling of steam:

Q1 = (mass of steam) × (specific heat capacity of steam) × (final temperature - initial temperature)

Q1 = (100.0 g) × (0.00203 kJ/(g°C)) × (100°C - 200°C)
Q1 = -20.3 kJ

Next, we need to calculate the heat lost during the phase change from steam to water at 100°C. The specific heat of vaporization for water is approximately 40.7 kJ/mol.

To find the moles of steam, we will use the molar mass of water, which is approximately 18 g/mol:

moles = (mass of steam) / (molar mass of water)
moles = (100.0 g) / (18 g/mol)
moles ≈ 5.56 mol

Now we can calculate the heat lost during the phase change:

Q2 = (moles of steam) × (heat of vaporization)
Q2 = (5.56 mol) × (40.7 kJ/mol)
Q2 ≈ -226.49 kJ

Finally, we need to calculate the heat lost during the cooling of water from 100°C to -30°C, using the same specific heat capacity as before:

Q3 = (mass of water) × (specific heat capacity of water) × (final temperature - initial temperature)

Q3 = (mass of water) × (4.18 J/(g°C)) × (100°C - (-30°C))
Q3 = (100.0 g) × (0.00418 kJ/(g°C)) × (100°C - (-30°C))
Q3 = -41.8 kJ

To find the total heat lost, we sum up all the individual heat losses:

Total heat lost = Q1 + Q2 + Q3
Total heat lost = -20.3 kJ + (-226.49 kJ) + (-41.8 kJ)
Total heat lost ≈ -288.59 kJ

Therefore, approximately 288.59 kJ of heat is lost when 100.0 g of steam at 200°C is cooled until it forms ice at -30.0°C.