Asked by Ryan
A quantity of ice at 0.0°C was added to 25.6 g of water at 21.0°C to give water at 0.0°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 · °C).
I don't understand how to set this problem up (what the formula is and where to plug in which qualities). Any help would be very much appreciated.
I don't understand how to set this problem up (what the formula is and where to plug in which qualities). Any help would be very much appreciated.
Answers
Answered by
DrBob222
How much heat must be removed from the 25.6 g H2O @ 21.0.0 C to move it to 0.0 C.
That will be
q = heat to be removed = mass x specific heat water x delta T.
q = 25.6 g x 4.18 x 21.0 = about 2000 J (but you need to do it exactly).
Then 2000 J x 1 g/heat fusion = grams ice.
You are given the heat of fusion in kJ/mol. You must change that to J/g so it will fit the remainder of the problem. It's approximately 334 J/g but you need to verify that. Post your work if you get stuck.
That will be
q = heat to be removed = mass x specific heat water x delta T.
q = 25.6 g x 4.18 x 21.0 = about 2000 J (but you need to do it exactly).
Then 2000 J x 1 g/heat fusion = grams ice.
You are given the heat of fusion in kJ/mol. You must change that to J/g so it will fit the remainder of the problem. It's approximately 334 J/g but you need to verify that. Post your work if you get stuck.
Answered by
Ryan
it worked perfectly, thank you so much
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