Given that the specific heat capacities of ice and steam are 2.06 J/g°C and 2.03 J/g°C, the molar heats of fusion and vaporization for water are 6.02 kJ/mol and 40.6 kJ/mol, respectively, and the specific heat capacity of water is 4.18 J/g°C, calculate the total quantity of heat evolved when 20.6 g of steam at 216°C is condensed, cooled, and frozen to ice at -50.°C.
The easiest way to do this is in parts. It's easier to understand.
q1 = heat to move steam from 216 to 100.
q1 = mass steam x specific heat steam x delta T.
q2 = heat to condense steam at 100 C to liquid water at 100 C.
q2 = mass steam x heat vaporization
q3 = heat to move liquid water from 100 C to 0 C.
q3 = mass x specific heat water x delta T.
q4 = heat to freeze liquid water at 0 C to ice at 0 C.
q4 = mass water x heat fusion.
q5 = heat to move ice from 0 C to -50 C.
q5 = mass x specific heat ice x delta T.
total heat evolved = q1 + q2 + q3 + q4 + q5 = ??
Watch the units. I would make sure everything was done in joules.
Another note: You must change grams to moles for the mass if using J/mole or change J/mole to J/g if you want to use mass in grams.
I worked it through and obtained 68.89 (the difference probably is rounding) kJ which I would round to 68.9 kJ (3 s.f. is all we are allowed) and your answer would be rounded to 68.9 kJ.
To calculate the total quantity of heat evolved when steam is condensed, cooled, and frozen to ice, we need to consider the following steps:
1. First, we need to calculate the heat released when the steam is condensed at constant pressure. The formula to calculate the heat released during condensation is:
Q1 = mass * molar heat of vaporization
For this, we use the given molar heat of vaporization for water, which is 40.6 kJ/mol. However, we need to convert the mass of steam from grams to moles, as the molar heat of vaporization is given per mole. The molar mass of water is approximately 18.02 g/mol.
moles of steam = mass of steam / molar mass of water
Next, we calculate the heat released during condensation:
Q1 = moles of steam * molar heat of vaporization
To get the value in joules, we need to convert the molar heat of vaporization from kJ/mol to J/mol:
molar heat of vaporization = 40.6 kJ/mol * 1000 J/kJ
2. After condensation, we need to cool the water from 100°C to 0°C. The formula to calculate the heat released during cooling is:
Q2 = mass * specific heat capacity of water
Here, the mass is the same as before, which is 20.6 g. The specific heat capacity of water is given as 4.18 J/g°C.
3. Finally, we need to freeze the water from 0°C to -50°C. The formula to calculate the heat released during freezing is:
Q3 = mass * molar heat of fusion
The molar heat of fusion for water is given as 6.02 kJ/mol. Again, we need to convert this to J/mol:
molar heat of fusion = 6.02 kJ/mol * 1000 J/kJ
4. To get the total heat released during the entire process, we sum up the individual heats:
Total heat released = Q1 + Q2 + Q3
Now, we can plug in the given values and calculate the result.