posted by Sam on .
If 118 g of ice at 0.0°C is added to 1.55 L of water at 88°C, what is the final temperature of the mixture?
heat absorbed by melting ice + heat to raise water from zero C + heat lost by 88 C water.= 0
[mass ice x heat fusion] + [mass melted ice x specific heat water x (Tfinal-Tinitial)] + [mass separate water x specific heat water x (Tfinal-Tinitial)] = 0.
Substitute the numbers and solve for Tfinal.
how do i get the mass of separate water, if it is in L?
1.55L is 1550 mL (1550 cc) and if the density of water is 1.00 g/mL, then the mass of the water is 1550 grams. Most problems of this nature usually state that the water is to be considered as having a density of 1 g/cc.
um...one more question, what do you mean by "mass of melted ice"? how is it determined?
Let's see now. If I had 1 g ice and it melted into water, it would form 1 g water. So grams ice and grams melted ice must be the same. right?