How much heat is released when 75.0 g of steam at 100.0 c is cooled to ice at -15.0c?

I'm using 6.01 kl/mol for fusion of water, 75.4 j/mol-c for heat capacity of water and 36.4 j/mol- c for molar heat capacity of ice.

To determine the amount of heat released when steam is cooled to ice, you can use the following equation:

Q = (m × c × ΔT)steam + Qfusion + (m × c × ΔT)ice

Where:
Q is the total heat released
m is the mass of the substance (in this case, steam and ice)
c is the specific heat capacity
ΔT is the change in temperature
Qfusion is the heat released during fusion (condensation or freezing)

Now let's calculate the heat released step by step.

1. First, we need to determine the amount of heat released when the steam cools from 100.0 C to 0.0 C. We'll use the heat capacity of water (75.4 J/mol°C) to find the heat released.

Q1 = (m × c × ΔT)steam
= (75.0 g / 18.0 g/mol) × (75.4 J/mol°C) × (0.0 C - 100.0 C)
= -25065 J

Note: We divide the mass by the molar mass of water (18.0 g/mol) to convert grams to moles.

2. Next, we need to determine the heat released during fusion. We'll use the value of 6.01 kJ/mol for the fusion of water.

Q2 = Qfusion
= (75.0 g / 18.0 g/mol) × (6.01 kJ/mol)
= -2511 J

Note: We convert kilojoules to joules by multiplying by 1000.

3. Finally, we need to determine the amount of heat released when the ice cools from 0.0 C to -15.0 C. We'll use the molar heat capacity of ice (36.4 J/mol°C) to find the heat released.

Q3 = (m × c × ΔT)ice
= (75.0 g / 18.0 g/mol) × (36.4 J/mol°C) × (-15.0 C - 0.0 C)
= -7567 J

Note: The change in temperature is calculated as -15.0 C - 0.0 C, as it is a decrease.

Now, we can calculate the total heat released:

Q = Q1 + Q2 + Q3
= -25065 J + (-2511 J) + (-7567 J)
= -35143 J

Therefore, the total heat released when 75.0 g of steam at 100.0 C is cooled to ice at -15.0 C is -35143 J.