How much energy (in kilojoules) is released when 13.3g of steam at 114.5∘C is condensed to give liquid water at 67.0∘C? The heat of vaporization of liquid water is 40.67 kJ/mol, and the molar heat capacity is 75.3 J/(K⋅mol) for the liquid and 33.6 J/(K⋅mol) for the vapor.

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To calculate the energy released when steam is condensed, we can follow these steps:

Step 1: Calculate the heat absorbed by the steam as it cools down from 114.5°C to 100°C.
Q1 = mass × specific heat capacity (liquid) × temperature change

The specific heat capacity of liquid water is given as 75.3 J/(K⋅mol). However, we need to convert the mass of steam from grams to moles before using this value.
To convert mass to moles, we use the molar mass of water (18.015 g/mol):
moles of steam = mass of steam / molar mass of water
moles of steam = 13.3 g / 18.015 g/mol

Now, we can calculate Q1:
Q1 = (moles of steam) × specific heat capacity (liquid) × temperature change
Q1 = (13.3 g / 18.015 g/mol) × 75.3 J/(K⋅mol) × (100°C - 114.5°C)

Step 2: Calculate the heat absorbed as the steam is condensed at 100°C.
Q2 = heat of vaporization × moles of steam

The heat of vaporization of liquid water is given as 40.67 kJ/mol. We need to convert this to joules before using the value in the calculation:
heat of vaporization = 40.67 kJ/mol × 1000 J/kJ

Now, we can calculate Q2:
Q2 = heat of vaporization × (moles of steam)

Step 3: Calculate the heat released as the liquid water cools down from 100°C to 67°C.
Q3 = mass × specific heat capacity (liquid) × temperature change

We can use the same specific heat capacity value as in Step 1.

Now, we can calculate Q3:
Q3 = mass × specific heat capacity (liquid) × temperature change
Q3 = 13.3 g × 75.3 J/(K⋅mol) × (100°C - 67°C)

Finally, we sum up all the calculated values to get the total energy released:
Total energy released = Q1 + Q2 + Q3

To solve this problem, we need to consider two parts: the energy required to cool the steam from 114.5∘C to 100∘C and then condense the steam at 100∘C to liquid water at 67.0∘C.

First, let's calculate the energy required to cool the steam from 114.5∘C to 100∘C.

DeltaT1 = (100∘C - 114.5∘C) = -14.5∘C

To calculate the energy required to cool the steam, we use the formula:

Energy1 = mass × specific heat capacity × ΔT1

The mass is given as 13.3g.
The specific heat capacity for the vapor phase is given as 33.6 J/(K⋅mol).

However, we need to convert the mass from grams to moles using the molar mass of water. The molar mass of water is approximately 18.015 g/mol.

moles = mass / molar mass = 13.3g / 18.015 g/mol

Now that we have the moles of water, we can calculate the number of kilojoules required to cool the steam:

Energy1 = (moles × specific heat capacity × ΔT1) / 1000

Next, let's calculate the energy required to condense the steam at 100∘C to liquid water at 67.0∘C.

The heat of vaporization of liquid water is given as 40.67 kJ/mol.

Energy2 = moles × heat of vaporization

Now, let's calculate the total energy released when the steam is condensed:

Total energy = Energy1 + Energy2

Finally, let's substitute the values into the equations and calculate the answer.

Do this in steps.

1. heat released in lowering steam from 114.5 to steam at 100 c.
q1 = mass steam x specific heat steam x (Tfinal-Tinitial)

q2 = heat released in changing steam at 100 C to liquid H2O at 100 C.
q2 = mass x heat vap

q3 = heat released in moving liquid H2O at 100 C to liquid H2O at 67 C

q3 = mass H2O x specific heat H2O x (Tfinal-Tinitial)

Total q = q1 + q2 + q3
NOTE that mass in in grams and specific heat and heat vap is in kJ/mol or J/mol.