Efficiency of engine is given by $$\eta = 1 - \frac{{{T_2}}}{{{T_1}}}$$
$${T_2} =$$  temperature of sink
$${T_1} =$$  temperature of source
$$\therefore \frac{{{T_2}}}{{{T_1}}} = 1 - \eta = 1 - \frac{1}{6} = \frac{5}{6}\,......\left( {\text{i}} \right)$$
In other case,
$$\eqalign{ & \frac{{{T_2} - 62}}{{{T_1}}} = 1 - \eta = 1 - \frac{2}{6} = \frac{2}{3}\,......\left( {{\text{ii}}} \right) \cr & {\text{or}}\,\,{T_2} - 62 = \frac{2}{3}{T_1} = \frac{2}{3} \times \frac{6}{5}{T_2}\,\,\left[ {{\text{Using}}\,{\text{Eq}}{\text{.}}\left( {\text{i}} \right)} \right] \cr & {\text{or }}\frac{1}{5}{T_2} = 62 \cr & \therefore {T_2} = 310\;K \cr & = 310 - {273^ \circ }C = {37^ \circ }C \cr} $$
Here, $${T_1} = \frac{6}{5}{T_2} = \frac{6}{5} \times 310$$
$$\eqalign{ & = 372\,K = 372 - 273 \cr & = {99^ \circ }C \cr} $$

$$\eqalign{ & {\text{Efficiency}}\,\,\eta = \frac{{{W_{{\text{output}}}}}}{{{\text{ }}Hea{t_{{\text{input}}}}}} = \frac{w}{{3w}} = \frac{1}{3} \cr & \eta = 1 - \frac{{{Q_2}}}{{{Q_1}}} = \frac{1}{3} \cr & \therefore \frac{{{Q_2}}}{{{Q_1}}} = \frac{2}{3} \cr} $$

Work done by the system in the cycle $$=$$ Area under $$P-V$$  curve and $$V$$-axis
$$\eqalign{ & = \frac{1}{2}\left( {2{P_0} - {P_0}} \right)\left( {2{V_0} - {V_0}} \right) + \left[ { - \left( {\frac{1}{2}} \right)\left( {3{P_0} - {P_0}} \right)\left( {2{V_0} - {V_0}} \right)} \right] \cr & = \frac{{{P_0}{V_0}}}{2} - \frac{{{P_0}{V_0}}}{2} = 0 \cr} $$

The relation between true scale and new scale of temperature is given by
$$\eqalign{ & {\left( {\frac{{t - LFP}}{{UFP - LFP}}} \right)_{{\text{true}}}} = {\left( {\frac{{t - LFP}}{{UFP - LFP}}} \right)_{{\text{faulty}}}} \cr & \Rightarrow \frac{{{{39}^ \circ }C - {0^ \circ }C}}{{{{100}^ \circ }C - {0^ \circ }C}} = \frac{{t - {{39}^ \circ }W}}{{{{239}^ \circ }W - {{39}^ \circ }W}} \cr & \Rightarrow t = {117^ \circ }W \cr} $$

For given cyclic process, $$\Delta U = 0 \Rightarrow Q = W$$
Also, $$W = - $$  area enclosed by the curve
$$\eqalign{ & = - AB \times AD \cr & = - \left( {2p - p} \right)\left( {3V - V} \right) \cr & = - p \times 2V \cr} $$
∴ Heat rejected $$ = 2\,pV$$

Here, work done is zero.
So, loss in kinetic energy = change in internal energy of gas
$$\eqalign{ & \frac{1}{2}m{v^2} = n{C_v}\,\Delta T \cr & = n\frac{R}{{\gamma - 1}}\,\Delta T \cr & \frac{1}{2}m{v^2} = \frac{m}{M}\frac{R}{{\gamma - 1}}\,\Delta T \cr & \therefore \,\,\Delta T = \frac{{M{v^2}\left( {\gamma - 1} \right)}}{{2R}}K \cr} $$

From the first law of thermodynamics
$$\eqalign{ & dQ = dU + dW \cr & {\text{Here}}\,dW = 0\,\left( {{\text{given}}} \right) \cr & \therefore dQ = dU \cr & {\text{Now}}\,{\text{since}}\,dQ < 0\,\left( {{\text{given}}} \right) \cr & \therefore dQ\,{\text{is}}\,{\text{negative}} \cr & \Rightarrow dU = - ve \Rightarrow dU\,{\text{decreases}}{\text{.}} \cr} $$
$$ \Rightarrow $$ Temperature decreases.

$$\eqalign{ & \Delta U = \Delta Q = mc\,\Delta T \cr & = 100 \times {10^{ - 3}} \times 4184\left( {50 - 30} \right) \cr & \approx 8.4\,k\,J \cr} $$

According to question, firstly gas expands from volume $$V$$ to $$3V$$ and after this volume is reduced from $$3V$$ to $$V$$ at constant pressure.
In isothermal expansion, $$p$$ - $$V$$ curve is rectangular hyperbola.

$$\eta = 1 - \frac{{{T_2}}}{{{T_1}}}$$
So for $$\eta $$ be high $${{T_1}}$$ must be high and $${{T_2}}$$ must be low.