$20 \ cal$ of heat is required to increase the temperature of a $15 \ g$ piece of $Al$ metal from $25^{\circ}C$ to $30^{\circ}C$. Calculate the heat capacity,specific heat capacity,and molar heat capacity of the $Al$ piece. (Given: Atomic mass of $Al = 27 \ g \ mol^{-1}$)

(N/A) Given: $q = 20 \ cal$,$\Delta T = 30^{\circ}C - 25^{\circ}C = 5^{\circ}C$,$m = 15 \ g$,$M = 27 \ g \ mol^{-1}$.
$1$. Heat capacity $(C)$: $C = \frac{q}{\Delta T} = \frac{20 \ cal}{5^{\circ}C} = 4.0 \ cal \ {}^{\circ}C^{-1}$.
$2$. Specific heat capacity $(c)$: $c = \frac{C}{m} = \frac{4.0 \ cal \ {}^{\circ}C^{-1}}{15 \ g} \approx 0.267 \ cal \ g^{-1} \ {}^{\circ}C^{-1}$.
$3$. Molar heat capacity $(C_m)$: $C_m = c \times M = 0.267 \ cal \ g^{-1} \ {}^{\circ}C^{-1} \times 27 \ g \ mol^{-1} \approx 7.21 \ cal \ mol^{-1} \ {}^{\circ}C^{-1}$.