A quantity of ice at 0.0 degrees C was added to 33.6 of water at 21.0 degree C to give water at 0.0 degrees C. How much ice was added? The heat of fusion of water is 6.01 kJ/mol and the specific heat is 4.18 J/(g * degrees C)
q = mass x specific heat x delta T?
q = mass x heat fusion?
I don't understand what to plug in
and where
Heat removed from water in going from 21.0 to zero C is
mass x specific heat x delta T.
mass = 33.6 g
specific heat is 4.184 J/g*K
delta T is 21.0 - 0.
heat added to ice to melt it is
mass x heat fusion.
mass is the unknown.
heat fusion is given.
Note: you need to work in the same units. The heat of fusion is quoted in the problem in kJ/mol but I quoted the specific heat of water in J/gram. One needs to be changed. I would suggest the 4.184 be changed.
ok so i did 33.6g (4.184 J/g)(21.0-0)
So how do i get J to kJ/mol
The specific heat of water in J/mol = 75.3. You can change 33.6 g water to mols by 33.6/18 = ??
Then (33.6/18)*75.3*21 = ?? Joules.
You can change to kJ by ??/1000 = xx.
Well, to convert from J (joules) to kJ (kilojoules), you need to divide by 1000 since there are 1000 joules in one kilojoule.
So, if you have the answer in joules, you can convert it to kilojoules by dividing by 1000.
To convert from J to kJ/mol, you need to consider the molar mass of water. The molar mass of water is approximately 18.015 g/mol.
You can use this conversion factor to convert the specific heat from J/g to J/mol:
1 J/g = (1 J/g) x (1 mol / 18.015 g) = 0.0555 J/mol
So, the specific heat of water can be expressed as 0.0555 J/mol.
Now you can calculate the heat removed from the water using the equation:
q = mass x specific heat x delta T
Where:
mass = 33.6 g
specific heat = 0.0555 J/mol
delta T = 21.0 - 0 = 21.0 degrees C
q = 33.6 g x 0.0555 J/mol x 21 = 38.4696 J
Since heat of fusion is given in kJ/mol, you need to convert the heat removed from the water to kJ by dividing by 1000:
38.4696 J / 1000 = 0.0384696 kJ
Now, to find the mass of ice added, you can use the equation:
q = mass x heat fusion
Where:
q = 0.0384696 kJ (converted heat removed from water)
heat fusion = 6.01 kJ/mol (given heat of fusion of water)
To calculate the mass of ice, rearrange the equation:
mass = q / heat fusion
mass = 0.0384696 kJ / 6.01 kJ/mol
mass ≈ 0.0064 mol
Since the molar mass of ice (H2O) is approximately 18.015 g/mol, you can find the mass of ice in grams by multiplying the moles of ice by the molar mass:
mass = 0.0064 mol x 18.015 g/mol
mass ≈ 0.1152 g
Therefore, approximately 0.1152 grams of ice was added.
To convert the specific heat from J/g to J/mol, you need to know the molar mass of water, which is 18.01528 g/mol.
To convert from J/g to J/mol, you can use the following conversion factor:
(1 g/mol)/(molar mass of water in g/mol)
So in this case, it would be:
(1 g/mol)/(18.01528 g/mol) = 0.0555 mol/g
To convert from J to kJ, you need to divide by 1000. Therefore, the final conversion factor is:
0.0555 mol/g * (1 kJ/1000 J) = 0.0000555 kJ/g
So the specific heat of water in kJ/mol*°C is 0.0000555 kJ/g*°C.
Now, you can plug this value into the equation:
q = mass x specific heat x delta T
q = 33.6 g x 0.0000555 kJ/g°C x (21.0 - 0)
Simplifying this expression will give you the heat removed from the water.