if 120 g of hot water at 85 degree celsius is poured into an insulated cup containing 200 g of ice at 0 degree celsius. how many grams of liquid will there be when the system reaches thermal equilibrium?

Answer 1

heat to melt x grams ice: x*Hfusion
heat absorbed by warm water: 120*cw*(tf-85)
heat absorbed by melted ice to get to final temperataure: x*cw*(tf-0)

the sum of the heats gained is zero. x is the amount of ice that melted.

xHf+( x)cw*(tf-0)+120(cw)(Tf-85)=0

first, see if all the ices melts, and check Tf.

200*80+200*1*tf+120*1*(tf-85)=0
tf=(120*85-16000 )/200=-29 which is impossible.

So all the ice did not melt, and the final temp is Tf=0

then
x*80+(200-x)Cw(Tf-0)+120*Cw*(Tf-85),but Tf=o if all the ice did not melt, or
80x+0=120*1*85
solve for x, or x=127.5 grams melted, so the water is 120+(200-127.5) g

check my thinking.