What is the final temperature and state of water when 8.50 kJ of energy is added to 10.0 g of ice at 0 degree Celsius.
Helpful information:
Cs (ice)= 2.1 J/g•C
Heat of vaporization (water)= 40.7kJ/mol
Ca (water)= 4.184 J/g•C
Heat of fusion (water)= 6.02 kJ/mol
Cs (steam) = 2.0 J/g• C
Energy used to melt ice at zero C to liquid water at zero C.
dq = mass ice x heat fusion = (10/18) x 6.02 kJ/mol = 3.34 kJ.
Energy remaining to be used is 8.50 kJ - 3.34 = 5.15 kJ.
Energy used to raise T of water at zero C to 100 C.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q = (10) x 4.184 x (100-0) = about 4184 J or 4.184 kJ.
How much is left now? That's 5.15 kJ - 4.18 kJ = about 0.98 kJ.
That can be used to vaporize some of the water at 100 C.
q = mass H2O x heat vap
0.97 kJ = mols H2O x 40.7 kJ/mol and solve for mols H2O, then convert to grams.
You need to go through all of these calculations again but you see you will have some H2O liquid and some water steam. Check all of these numbers.
To find the final temperature and state of water, we need to consider the heat transfer involved in three steps:
1. Heat required to melt the ice at 0 degrees Celsius to liquid water at 0 degrees Celsius.
2. Heat required to heat the liquid water from 0 degrees Celsius to the boiling point.
3. Heat required to convert the boiling water to steam at the boiling point.
Let's calculate the heat required for each step:
1. Heat required to melt the ice at 0 degrees Celsius to liquid water at 0 degrees Celsius:
- We have 10.0 g of ice, and the heat of fusion for water is 6.02 kJ/mol.
- First, convert the mass of ice to moles: 10.0 g / (18.0 g/mol) = 0.556 mol.
- Calculate the heat required to melt the ice: 0.556 mol x 6.02 kJ/mol = 3.35 kJ.
2. Heat required to heat the liquid water from 0 degrees Celsius to the boiling point:
- The specific heat capacity of liquid water (Ca) is 4.184 J/g•C.
- The change in temperature is from 0 degrees Celsius to the boiling point (100 degrees Celsius).
- We need to calculate the heat required for this temperature change using the formula: Q = m x Ca x ΔT, where Q is the heat, m is the mass, Ca is the specific heat capacity, and ΔT is the temperature change.
- Calculate the heat required: 10.0 g x 4.184 J/g•C x (100 - 0) = 4184 J = 4.184 kJ.
3. Heat required to convert the boiling water to steam at the boiling point:
- The heat of vaporization for water is 40.7 kJ/mol.
- The mass of water is still 10.0 g, and we need to convert it to moles as before: 10.0 g / (18.0 g/mol) = 0.556 mol.
- Calculate the heat required to convert the water to steam: 0.556 mol x 40.7 kJ/mol = 22.63 kJ.
Now let's add up the heat for each step:
- Heat from step 1: 3.35 kJ
- Heat from step 2: 4.184 kJ
- Heat from step 3: 22.63 kJ
Total heat added = 3.35 kJ + 4.184 kJ + 22.63 kJ = 30.164 kJ
Since we have added 8.50 kJ of energy, the remaining heat is: 8.50 kJ - 30.164 kJ = -21.664 kJ.
Since we have excess energy, the water will reach its boiling point of 100 degrees Celsius, and the excess energy will be converted to steam.
Therefore, the final temperature is 100 degrees Celsius, and the final state is steam.