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.
a 35 g ice cube at 0.0 C is added to 110 g of water in a 62 g aluminium cup. The cup and the water have an initial temperature of 23 C . Find the equilibrium temperature of this system? in Celsius ,, I've been trying to solve this for 2 hours now , and I still can't get the answer ,, help please ?
3 answers
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.
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
2 answers