Answer
$1700\frac{J}{kg\cdot C^{\circ}}$
Work Step by Step
Use equation 14–2.
$$Q=mc\Delta T$$
$$135\times10^3J =(4.1kg)c (37.2^{\circ}C-18.0^{\circ}C) $$
Solve for $c=1715\frac{J}{kg\cdot C^{\circ}}\approx1700\frac{J}{kg\cdot C^{\circ}}$
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