Conceptual Physics (12th Edition)

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

Chapter 17 - Think and Solve: 38

Answer

320 calories. It is less.

Work Step by Step

Changing 1 gram of $-273^{\circ}C$ ice to $0^{\circ}C$ ice requires (273)(0.5) = 140 calories. Melting that gram of $0^{\circ}C$ ice to $0^{\circ}C$ of water requires 80 calories. Raising the temperature of 1 gram of $0^{\circ}C$ water to $100^{\circ}C$ boiling water requires 100 calories. The subtotal is 320 calories. This is to be compared with the energy needed to change 1 gram of $100^{\circ}C$ water to steam at the same temperature, which is 540 calories. The latter process, perhaps surprisingly, takes more energy than it took to bring the water all the way from absolute zero to $100^{\circ}C$!
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