Calculate the heat released when 42.0 g of water at 35.0 degrees Celsius is converted to ice at -5.0 degrees Celsius. The specific heat of ice is 2.03 J/(g C), the molar heat of fusion of ice is 6010 J/mol, and the specific heat of water is 4.18 J/(g C)
(Hint: Be sure to consider the sign of the heat of fusion vs. heat of freezing.)
I obviously don't know how to do this since my two out three attempts were wrong. can someone please explain so I can figure out what i'm doing wrong?
Why didn't you show your work so we could figures out what you were doing wrong?
You want to add the heat removed in changing T from 35 to zero, to the heat removed @ zero C to convert to ice, to the heat removed in changing T from zero C to -5.
q=mass*Specific heat of liquid*change in temp
q=42*4.18*35=6144.6
q=mass* specific heat of ice*change in temp
q=42*2.03*5=426.3
q=heat of fusion of ice*moles
q=6010*2.3=14023
Then add all three together and get 20593.9 Joules
Your numbers look ok. I suspect the problem is that you are reporting too many significant figures.
For example, that first one of 5144.6 is allowed only 3 places. Also, shouldn't you have a - sign for the total. Each of the three are exothermic; i.e.,, the first one is mass x sp. h. x (Tfinal-Tinitial) = Tf is 0 and Ti is 35 so that gives you a negative sign. Frankly, I think when the problem says realeased that you need not consider the - sign because you already know it's released. However, it appears obvious from the problem that the author expect the sign to be included.
To solve this problem, we need to consider two steps: heating the water from 35.0°C to 0°C and then freezing the water at 0°C to ice at -5.0°C.
Step 1: Heating the water from 35.0°C to 0°C
To calculate the heat released during this step, we'll use the formula:
q1 = m * C * ΔT
where:
q1 = heat released
m = mass of water (42.0 g)
C = specific heat of water (4.18 J/(g°C))
ΔT = change in temperature (0°C - 35.0°C = -35.0°C)
Calculating q1:
q1 = 42.0 g * 4.18 J/(g°C) * (-35.0°C)
q1 = -62,286 J (Note: The negative sign indicates heat release)
Step 2: Freezing the water from 0°C to -5.0°C
To calculate the heat released during this step, we'll use the formula:
q2 = m * ΔHf
where:
q2 = heat released
m = mass of water (42.0 g)
ΔHf = molar heat of fusion of ice (6010 J/mol)
First, we need to calculate the number of moles of water:
moles = mass / molar mass
The molar mass of water (H2O) is approximately 18.015 g/mol. Substituting the values:
moles = 42.0 g / 18.015 g/mol
moles = 2.33 mol (rounded to two decimal places)
Now, we can calculate q2:
q2 = 2.33 mol * 6010 J/mol
q2 = 14,033.3 J (rounded to one decimal place)
Finally, to get the total heat released, we sum q1 and q2:
Total heat released = q1 + q2
Total heat released = -62,286 J + 14,033.3 J
Total heat released = -48,252.7 J (rounded to one decimal place)
Therefore, the total heat released when 42.0 g of water at 35.0°C is converted to ice at -5.0°C is approximately -48,252.7 J. The negative sign indicates the release of heat during the process.