Answer
$Q = 82 cal$
Work Step by Step
Given:
c = 0.20 + 0.14$T$ + 0.023$T^{2}$ (with c in $\frac{cal}{g*K}$ and T in $^{\circ}$C)
m = 2.0 g
Ti = 5.0$^{\circ}$C
Tf = 15$^{\circ}$C
Since c is variable, we must integrate first, from 5 to 15$^{\circ}$C:
c = $\int$(0.20 + 0.14$T$ + 0.023$T^{2}$)dT
Integrating:
c = (0.20(15) + $\frac{0.14(15)^2}{2}$ + $\frac{0.023(15)^{3}}{3}$) - (0.20(5) + $\frac{0.14(5)^2}{2}$ + $\frac{0.023(5)^{3}}{3}$) $\approx$ 40.9167 $\frac{cal}{g}$
Plugging into Q = mc: (Note the units on Q end up as $\frac{cal}{g}$, so:
Q = (2.0 g)(40.9167 $\frac{cal}{g}$) $\approx$ 82 cal
I use $\approx$ as I rounded to two significant figures.
