Answer
$2831.378 J $
Work Step by Step
Using the equation $q=mc\Delta T$, we can solve for q which in this case is the heat that is required to heat the aluminum sample from $72^{\circ}F $ to $145^{\circ}F$. The change in temperature can be calculated by taking the final temperature and subtracting it from the initial. Therefore, its $145-72 = 73^{\circ}F$. The heat capacity ("c") of aluminum is $0.902 \frac{J}{g\times^{\circ}C}$
Therefore, $q = 43 g \times 0.902 \frac{J}{g\times^{\circ}C}\times73= 2831.378 J $
