Answer
$ Q_H = 7196 J $
Work Step by Step
To find the net energy that enters the system as heat, first we need to find the heat as constant volume $Q_V$ and the heat in isothermal process, $Q_{isothermal}$
$Q_V = nC_V\Delta T$
For monatomic gas, $C_V = \frac{3}{2}R$
and R = 8.314 J/mol.K. So
$Q_V = n \frac{3}{2}R\Delta T$
$Q_V = (1 mol) \frac{3}{2}(8.31 J/mol.K)(600K - 300K)$
$Q_V = 3740 J$
Next, find the $Q_{isothermal}$
$Q_{isothermal} = nRT_Hln(2)$
$Q_{isothermal} = (1 mol)(8.31 J/mol.K)(600K)(ln2)$
$Q_{isothermal} = 3456 J$
So, total of net energy that enters the system as heat is
$Q_H = Q_V+ Q_{isothermal}$
$ Q_H = 3740 J + 3456 J $
$ Q_H = 7196 J $
