Answer
$Q = 1200~cal$
Work Step by Step
We can find $C_p$:
$C_p = C_v+R$
$C_p = (6.00~cal/mol~K)+(2.0~cal/mol~K)$
$C_p = 8.00~cal/mol~K$
We can find $Q$:
$C_p = \frac{Q}{n~\Delta T}$
$Q = C_p~n~\Delta T$
$Q = (8.00~cal/mol~K)(3.00~mol)(50.0~K)$
$Q = 1200~cal$
