#### Answer

The final temperature of the water will be the same as that of the very large block of ice, $0^{\circ}C$.

#### Work Step by Step

We start by finding the heat released when 50 g of $80^{\circ}C$ water cools to 50 g of $0^{\circ}C$ water. This is discussed in the Check Point on page 331, item 2, using 50 grams instead of 1 gram.
50 g of $80^{\circ}C$ water cooling to $0^{\circ}C$ gives up 4000 calories.
The 4000 calories, is used to melt ice. We know that 1 g of $0^{\circ}C$ ice requires 80 calories to melt to 1 g of $0^{\circ}C$ water, so we can melt (4000/80) = 50 grams of ice, the same mass of $80^{\circ}C$ water that we started with.