Answer
$2^{\circ}C$
Work Step by Step
Use equation 14–2.
$$Q=mc\Delta T $$
$$\Delta T= \frac{Q}{mc}=\frac{(0.80)(200kcal/h)(0.75h)}{(70 kg)(0.83kcal/(kg\cdot C^{\circ}))}=2^{\circ}C$$
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