Conceptual Physics (12th Edition)

Published by Addison-Wesley
ISBN 10: 0321909100
ISBN 13: 978-0-32190-910-7

Chapter 17 - Think and Solve: 41


The final temperature of the iron will be the same as that of the very large block of ice, $0^{\circ}C$ . We start by finding the heat released when 50 g of $80^{\circ}C$ iron cools to 50 g of $0^{\circ}C$ iron. $$Q = cm \Delta T = (0.11 \frac{cal}{g^{\circ}C})(50 g)(80^{\circ}C ) = 440 cal$$ The 440 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 (440/80) = 5.5 grams of ice. The $80^{\circ}C$ iron can melt less ice than an equal mass of $80^{\circ}C$ water could, which is due to the iron's lower specific heat capacity.
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