Answer
$\Delta K = 450~cal$
Work Step by Step
We can find the change $\Delta K$ in the total translational kinetic energy:
$\Delta K = 1.5~k~\Delta T~N$
$\Delta K = (1.5)~(1.38\times 10^{-23}~J/K)~(50.0~K)~(3.00~mol)(6.02\times 10^{23}~mol^{-1})$
$\Delta K = 1869~J$
$\Delta K = (1869~J)(\frac{1~cal}{4.184~J})$
$\Delta K = 450~cal$
