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Stuck on Change of Phase question (heat)

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stuck on Change of Phase problem

How much ice at 0 degrees Celsius must be mixed with 50.0 g of water at 75.0 degrees Celsius to give a final water water temp. of 20 degrees Celsius?


ive been trying to work this out for the past hour, but having trouble...any help would be appreciated.
 
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How much ice at 0 degrees Celsius must be mixed with 50.0 g of water at 75.0 degrees Celsius to give a final water water temp. of 20 degrees Celsius?


ive been trying to work this out for the past hour, but having trouble...any help would be appreciated.
 
Gokul43201
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Show us what you've tried, and we'll help you out from there.
 
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You need to find two equations.

1. the energy required raise x grams of water by 1 degree
2. the energy required to melt x grams of ice

The energy in the system has to balance. Energy in from cooling the hot water must equal the energy out melting the ice and warming the resultant cold water.

Come back if you have the equations but are still stuck.
 
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ok, well i know ice at 0 degrees C ---->water at 0 deg. C , so you use the vaporization law Q = mLv so 'm' is the variable in this case, since im trying to find how much of the ice was needed, so the formula is :

Q = 'm'(2.26 x 10^6)

and then water at 0 deg. C ----> water at 75 deg. C

so i used Q = mC/\T

so plugging in the data, it turns into Q = .05kg(4,186)(20 deg. - 75 deg)



so i have : m(2.26x10^6) + .06(4,186)(20-75)

I don't even know if I'm applying the right principles to the problem, but thats what I THINK so far,...any corrections and continuation help would be appreciated
 
17
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i know the specific heat of water is 4,186..so its 4,186 J/kg C to raise x grams of water by one degree


and the energy required to melt ice is the latent heat of fusion..which is Q = MLf

i just dont know how to appy this and use it.
 
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
6,987
14
ok, well i know ice at 0 degrees C ---->water at 0 deg. C , so you use the vaporization law Q = mLv so 'm' is the variable in this case, since im trying to find how much of the ice was needed, so the formula is :

Q = 'm'(2.26 x 10^6)
Correct, so far. Don't forget the units.

and then water at 0 deg. C ----> water at 75 deg. C
Which water is this, the water from the ice, or the pre-existing water? Does the water from the ice warm up all the way to 75*C?
so i used Q = mC/\T

so plugging in the data, it turns into Q = .05kg(4,186)(20 deg. - 75 deg)
This is now different water that you are talking about here compared to above. This is the energy released by the existing warm water, as it cools to 20*C.

so i have : m(2.26x10^6) + .06(4,186)(20-75)
This part is not correct.

The logic you want to use is based on the principle of energy conservation for this system: since the total energy of the system is constant, the heat lost by the cooling water (which you calculated above) must be equal to the total heat gained by (i) the melting of ice and (ii) the subsequent warming of this water up to 20*C.


PS: Please do not doublepost. Also, from the next time on, use the Homework & Coursework subforums for questions like this.
 

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