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Fusion related problem (help please, chem final is tomorrow!)

  • Thread starter mathzeroh
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ok my final exam for chem is tomorrow and i need help badly on this practice problem! here it is:

3. When 9.250 kJ of heat is added to 20.0 g of ice at 0.0 degrees Celcius, what is the final temperature of the water? The heat of fusion for water is 335 J/g.

I don't know how on earth to connect the fact that i was given the "fusion" (fusion - huh???) of water with the other stuff. I know that q=(c)(m)(delta T) but how do i connect all this together?

this is what i did (and failed misrebaly):

q=(c)(m)(delta T); m=20.0 g; T initial=0.0 degrees Celcius; q=9.250 kJ (which is 9250 J if i wanted to work with the same units)

so:

9250 J=[4.184J/(g*degree Celcius)](20.0 g)(T final - 0.0 deg. Celcius)
9250 J=(83.68)(T final)
111 deg. Celcius=T final


WRONG!!

HELP PLEASE! :cry: :uhh: :confused:
 

OlderDan

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Heat of fusion, often called "latent heat of fusion" is the amount of heat needed to melt ice without changing its temperature, or the amount of heat you need to remove from water at freezing temperature to change it ("fuse" it) into ice.
 
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OlderDan said:
Heat of fusion, often called "latent heat of fusion" is the amount of heat needed to melt ice without changing its temperature, or the amount of heat you need to remove from water at freezing temperature to change it ("fuse" it) into ice.
Oh, hey thanks a lot.

But how do I relate that bit of information to the problem?
 
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where is everybody?
 

OlderDan

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mathzeroh said:
Oh, hey thanks a lot.

But how do I relate that bit of information to the problem?
You figure out how much heat is needed just to melt the ice you started with before you calculate the temperature change of the water resulting from the rest of the heat added.
 

HallsofIvy

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1. Determine how much of the heat is required to melt the ice- that is the heat of fusion times the mass of the ice.
2. Subtract that from the original 9.25 J to see how much heat is left to raise the temperature of the water.
3. Calculate how many degrees that much heat will raise the water- that is the number of degrees one J of heat will raise 1 g of water times the number of J left times the mass of water.
 
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ans

3. When 9.250 kJ of heat is added to 20.0 g of ice at 0.0 degrees Celcius, what is the final temperature of the water? The heat of fusion for water is 335 J/g.

9.250kJ = 9250J
energy required to melt ice = 335 * 20 = 6700J
energy left after ice has melted = 9250 - 6700 = 2550J

taking specific heat capacity of water to be 4.2 J g^-1 degreeC^-1

therefore heat capacity for 20g of water is 84 J C^-1

increase in temperature = 2550 / 84 = 30.6 degree C

therefore final temperature = 0.0(initial temp) + 30.6 = 30.6 degree C
 

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