For cyclic process ;
$$\eqalign{ & {Q_{{\text{cyclic}}}} = {W_{AB}} + {W_{BC}} + {W_{CA}} \cr & = 10\,J + 0 + {W_{CA}} \cr & = 5\,J \cr & \Rightarrow \,\,{W_{CA}} = - 5\,J \cr} $$

Change in entropy is given by
$$\Delta S = \frac{{ml}}{T} = \frac{{1000 \times 80}}{{273}} = 293\,cal\,{K^{ - 1}}$$

$$\eqalign{ & {Q_1} = {T_0}\;{S_0} + \frac{1}{2}\;{T_0}\;{S_0} = \frac{3}{2}\;{T_0}\;{S_0} \cr & {Q_2} = {T_0}\left( {2\;{S_0} - {S_0}} \right) = {T_0}\;{S_0}\,\,{\text{and}}\,\,{Q_3} = 0 \cr & \eta = \frac{W}{{{Q_1}}} = \frac{{{Q_1} - {Q_2}}}{{{Q_1}}} \cr & = 1 - \frac{{{Q_2}}}{{{Q_1}}} = 1 - \frac{{{T_0}\;{S_0}}}{{\frac{3}{2}\;{T_0}\;{S_0}}} = \frac{1}{3} \cr} $$

The efficiency of heat engine is $$\eta = 1 - \frac{{{T_2}}}{{{T_1}}}$$
or $$\frac{W}{{{Q_1}}} = 1 - \frac{{{T_2}}}{{{T_1}}}$$
$${{T_2}} =$$  Temperature of sink
$${{T_1}} =$$  Temperature of source
$$W =$$  Work done
Given, $${{Q_1}} =$$  heat absorbed from the source $$=6\,kcal$$
$${T_1} = 227 + 273 = 500\,K$$
and $${T_2} = 127 + 273 = 400\,K$$
Hence, $$\frac{W}{6} = 1 - \frac{{400}}{{500}}$$
or $$\frac{W}{6} = \frac{{100}}{{500}}\,\,{\text{or}}\,\,W = 1.2\,kcal$$
Thus, amount of heat converted into work is $$1.2\,kcal.$$

In the first-case adiabatic change,
$$\Delta Q = 0,\Delta W = - 35\,J$$
From $${1^{st}}$$  law of thermodynamics,
$$\eqalign{ & \Delta Q = \Delta U + \Delta W, \cr & {\text{or}}\,\,0 = \Delta U - 35 \cr & \therefore \Delta U = 35\,J \cr} $$
In the second case
$$\eqalign{ & \Delta Q = 12cal = 12 \times 4.2\;J = 50.4\,J \cr & \Delta W = \Delta Q - \Delta U = 50.4 - 35 = 15.4\,J \cr} $$

Internal energy and entropy are state function, they do not depend upon path taken.

From the first law of thermodynamics
$$dQ = dU + dW$$
Here $$dW = 0$$  (given)
∴ $$dQ = dU$$
Now since $$dQ < 0$$  (given)
∴ $$dQ$$  is negative
⇒ $$dU = - ve$$
⇒ $$dU$$  decreases.
⇒ Temperature decreases.

According to first law of thermodynamics
$$\eqalign{ & \Delta Q = \Delta U + \Delta W \cr & \Delta U = \Delta Q - \Delta W \cr & \Delta Q = 35\,J,\Delta W = - 15\,J \cr & \therefore \Delta U = 35\,J - \left( { - 15\,J} \right) = 50\,J \cr} $$

For a diatomic gas,

$${C_V} = \frac{5}{2}R$$
The change in internal energy of gas in the transition from $$A$$ to $$B$$ is
$$\eqalign{ & \Delta U = n{C_V}dT \cr & = n\left( {\frac{{5R}}{2}} \right)\left( {{T_B} - {T_A}} \right) \cr & = nR\frac{5}{2}\left( {\frac{{{p_B}{V_B}}}{{nR}} - \frac{{{p_A}{V_A}}}{{nR}}} \right) \cr & = \frac{5}{2}\left( {2 \times {{10}^3} \times 6 - 5 \times {{10}^3} \times 4} \right) \cr & = \frac{5}{2} \times \left( { - 8 \times {{10}^3}} \right) \cr & = - \frac{{4 \times {{10}^4}}}{2} = - 20\,kJ \cr} $$

Work done by gas in going isothermally from state $$A$$ to $$B$$ is
$$\Delta {W_{AB}} = nRT\ln \frac{{{P_A}}}{{{P_B}}} = nRT\ln 2\,.....\left( {\text{i}} \right)$$
Work done by gas in going isothermally from state $$B$$ to $$C$$ is
$$\Delta {W_{BC}} = nRT\ln \frac{{{P_B}}}{{{P_C}}} = nRT\frac{{{P_0}}}{{2{P_C}}}\,.....\left( {{\text{ii}}} \right)$$
It is given that $$\Delta {W_{BC}} = 2\Delta {W_{AB}}$$
$$\eqalign{ & \ln \frac{{{P_0}}}{{2{P_C}}} = \ln {\left( 2 \right)^2} \cr & \therefore {P_C} = \frac{{{P_0}}}{8} \cr} $$