How much energy (in kJ) is released when 16.4 g of steam at 121.5 degrees C is condensed to give liquid water at 60.5 degrees 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.
q1 = energy released to convert steam atg 121.5 to steam at 100 C.
q1 = mass steam x specific heat steam x (100-121.5) = ?
q2 = energy released to condense steam at 100 to liquid water at 100 C.
q2 = mass steam x heat vaporization
q3 = energy released in cooling liquid water from 100 C to 60.5 C.
q3 = mass water x specific heat water x (60.4-100) = ?
Total energy released is q1 + q2 + q3.
To determine the energy released when steam condenses to liquid water, we need to consider the following steps:
Step 1: Calculate the amount of heat required to cool the steam from 121.5°C to 100°C.
Step 2: Calculate the heat released when the steam condenses at 100°C.
Step 3: Calculate the heat released when the liquid water is further cooled from 100°C to 60.5°C.
Let's go step by step:
Step 1: Calculate the amount of heat required to cool the steam from 121.5°C to 100°C.
The molar heat capacity for the vapor (steam) is given as 33.6 J/(K*mol).
To calculate the heat required, we can use the formula:
q = m * C * ΔT
where:
q = heat
m = mass of the substance
C = molar heat capacity
ΔT = change in temperature
In this case, we need to calculate the heat required to cool the steam from 121.5°C to 100°C. The mass of steam can be determined using its molar mass.
Molar mass of H2O = 18.01528 g/mol
Therefore, the number of moles of steam (n) can be calculated as:
n = mass / molar mass = 16.4 g / 18.01528 g/mol
Next, we can calculate the heat required to cool the steam using the formula:
q1 = n * C * ΔT1
where:
q1 = heat required to cool the steam
C = molar heat capacity for the vapor (steam)
ΔT1 = change in temperature (121.5°C - 100°C)
Plugging in the values:
q1 = n * C * ΔT1
= (16.4 g / 18.01528 g/mol) * (33.6 J/(K*mol)) * (121.5°C - 100°C)
Step 2: Calculate the heat released when the steam condenses at 100°C.
The heat of vaporization of liquid water is given as 40.67 kJ/mol.
We know that 1 mole of H2O vaporizes to form 1 mole of H2O vapor. So, the heat released when steam condenses can be calculated as:
q2 = n * Heat of vaporization
where:
q2 = heat released when steam condenses
Plugging in the values:
q2 = (16.4 g / 18.01528 g/mol) * (40.67 kJ/mol) * 1000 J/kJ
Step 3: Calculate the heat released when the liquid water is further cooled from 100°C to 60.5°C.
The molar heat capacity for liquid water is given as 75.3 J/(K*mol).
We can use the formula to calculate the heat released:
q3 = n * C * ΔT3
where:
q3 = heat released when liquid water is further cooled
C = molar heat capacity for liquid water
ΔT3 = change in temperature (100°C - 60.5°C)
Plugging in the values:
q3 = (16.4 g / 18.01528 g/mol) * (75.3 J/(K*mol)) * (100°C - 60.5°C)
Finally, to get the total energy released, we sum up all the calculated values:
Total energy released = q1 + q2 + q3
Now you can plug in the given values into the equations and perform the necessary calculations to find the answer.