Answer
$2^{\circ}C$
Work Step by Step
Use equation 14–2.
$$Q=mc\Delta T $$
$$\Delta T= \frac{Q}{mc}=\frac{(0.80)(200kcal/h)(0.75h)}{(70 kg)(0.83kcal/(kg\cdot C^{\circ}))}=2^{\circ}C$$
Physics: Principles with Applications (7th Edition)
by Giancoli, Douglas C.
Published by
Pearson
ISBN 10:
0-32162-592-7
ISBN 13:
978-0-32162-592-2
Chapter 14 - Heat - General Problems - Page 410: 51
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