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1. I cant seem to do this question at all

0.5L of water starts at 20 degrees c, 1.35 x10^6J of heat energy are put into to it, what is the final temp? (its 152 degrees c, but i cant get it)

c for liquid water is = 4.19 x 10^3Jk-1kg-1

and steam is = 2.26 x10^6

Thanks, rep will be given.
2. "rep will be given" hahahaha, please, no-one do it for rep but do do it...
3. Use dq = mcdT

dq = change in heat energy
m = mass of water (1g per ml for water)
c = heat capacity
dT = final temp - initial temp
4. (Original post by EierVonSatan)
Use dq = mcdT

dq = change in heat energy
m = mass of water (1g per ml for water)
c = heat capacity
dT = final temp - initial temp
We havn't done calculus yet.
5. (Original post by Carlo08)
We havn't done calculus yet.
I'm just using d for delta (change in)
6. well I'm presuming he used the dq=mcdT and so I presume you've forgotten about latent heat capacity.... could be horribly wrong here though....
7. Yeah i used Q=mcdt with dt as the subject and it doesn't work, you have to take into account the change of state and so use the latent heat capacity for steam but how...
8. you just need to take away the latent heat x mass away from dQ, but I get the answer to be 125 not 152

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Updated: November 11, 2008
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