# Chapter 17 - Think and Solve: 39

80 grams of ice.

#### Work Step by Step

We start by finding the heat released when 10 g of $100^{\circ}C$ steam changes to 10 g of $0^{\circ}C$ water. This is discussed in the Check Point on page 331, items 1 and 2, using 10 grams instead of 1 gram. 10 g of steam condensing to 10 g of boiling water at $100^{\circ}C$ gives up 5400 calories. 10 g of $100^{\circ}C$ water cooling to $0^{\circ}C$ gives up 1000 calories. The total, 6400 calories, is used to melt ice. Since 1 g of $0^{\circ}C$ ice requires 80 calories to melt to 1 g of $0^{\circ}C$ water, we can melt (6400/80) = 80 grams of ice.

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