c
As initial and final points are same so
$\Delta {U_{ABC}} = \Delta {U_{AC}}$
$AB$ is isochoric process.
$\Delta {W_{AB}} = 0$
$\Delta {Q_{AB}} = \Delta {U_{AB}} = 400\,J$
$BC$ is isobaric process.
$\Delta {Q_{BC}} = \Delta {U_{BC}} + \Delta {W_{BC}}$
$100 = \Delta {U_{BC}} + 6 \times {10^4}\left( {4 \times {{10}^{ - 3}} - 2 \times {{10}^{ - 3}}} \right)$
$100 = \Delta {U_{BC}} + 12 \times 10$
$\Delta {U_{BC}} = 100 - 120 = - 20\,J$
$As,\,\Delta {U_{ABC}} = \Delta {U_{AC}}$
$\Delta {U_{AB}} + \Delta {U_{BC}} = \Delta {Q_{AC}} - \Delta {W_{AC}}$
$400 - 20 = \Delta {Q_{AC}} - (2 \times {10^4} \times 2 \times {10^{ - 3}} + \frac{1}{2} \times $
$2 \times {10^{ - 3}} \times 4 \times {10^4})$
$\Delta {Q_{AC}} = 460J$
