Posted by AfterLife on Saturday, August 20, 2011 at 1:36pm.
The sum of the heats gained equals zero (yes, the substance losing heat is negative).
sum heats gained=0
35*Hfice+62(Calum)*(Tf-0)+35(Cwater)(Tf-0)+ 110*Cwater(Tf-23)=0
solve for Tf.
Assume all of the ice melts. (If it doesn't you will find out later when you compute the final temperature).
Look up the specific heat of aluminum. It is 0.22 cal/C gm
Assume the heat lost by the water and aluminum when being lowered to final temperature T equals the heat gained by the ice when melting and rising to the same temperature T. Solve for T .
( 110*1.0 + 62*0.22)*(23 - T) = 35*(80 + 1.0*T)
The 1.0 numbers are specific heats of water and 80 cal/g is the latent heat of fusion of ice
If you get a negative T (in C) , all of the ice did not melt.
Thanx I will try it out
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