Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 16 - Exercises and Problems - Page 310: 45


$ -10^{\circ}C$ $263.15 K$

Work Step by Step

We use the following equation to find: $\frac{h}{h_3} = \frac{T}{T_3}$ $\frac{57.8}{60} = \frac{T}{273.16}$ $ T = -10^{\circ}C$ This is the same as 263.15 degrees Kelvin.
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