Answer
Please see the work below.
Work Step by Step
We know that
$\Delta T_F=(\frac{9}{5})\Delta T_C$
We plug in the known values to obtain:
$\Delta T_F=(\frac{9}{5})(10)$
$\Delta T_F=18F^{\circ}$
Essential University Physics: Volume 1 (3rd Edition)
by Wolfson, Richard
Published by
Pearson
ISBN 10:
0321993721
ISBN 13:
978-0-32199-372-4
Chapter 16 - Section 16.4 - Exercises and Problems - Page 299: 18
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