The amount of heat which the water gives up in cooling to 0∘C is not enough to melt all the ice, since each gramme of 0∘ water will give up 50 cal in cooling, while heating 1 g of ice to 0∘C and melting it requires that 0⋅5 X 40 + 80 = 100 cal be expended (the thermal capacity of ice is 0⋅5 cal/deg).
On the other hand, the amount of heat which the ice will take in heating to 0∘ is not enough to freeze all the water, since each gramme of ice will absorb 20 cal in doing this, while for the water to be turned to ice, each gramme of water must be made to give up 50 + 80 = 130 cal. Thus both water and ice will remain in the calorimeter, and therefore the final temperature of the mixture will be 0∘C.