Answer
$\Delta E_{int}=4.99KJ$
Work Step by Step
We know that;
$\Delta E_{int}=nC_v\Delta T$
and $C_v=\frac{5}{2}R$
Thus, $\Delta E_{int}=n(\frac{5}{2}R)\Delta T$.
We plug in the known values to obtain:
$\Delta E_{int}=\frac{5}{2}(4.00)(8.31)(60.0)=4.99\times 10^3=4.99kJ$
