Answer:
[tex]T_f=328.1 K[/tex]
Explanation:
To determine the final temperature of the sample, we use the specific heat formular as follows:
[tex]Q=mCp(T_f-T_0)\\T_f=\frac{Q}{mCp} +T_{0}\\mol Al= \frac{15g}{26.98gmol^{-1}}=0.555 mol\\T_f=\frac{245.0J}{0.555mol\times24.35Jmol^{-1}K^{-1}} +310.0K\\T_f=328.1 K[/tex]
Finally, the temperature of the aluminium sample has raised 18 K.
