An ideal Carnot engine whose efficiency is $50 \%$ receives heat at $500 \ K$. If the efficiency is to be $60 \%$, then the intake temperature for the same exhaust temperature is: (in $K$)

Work done by a Carnot engine operating between temperatures $127^{\circ}C$ and $27^{\circ}C$ is $2\,kJ$. The amount of heat transferred to the engine by the source is $........\,kJ$.

Two Carnot engines $A$ and $B$ operate in series such that engine $A$ absorbs heat $Q_1$ at temperature $T_1$ and rejects heat $Q$ to a sink at temperature $T$. Engine $B$ absorbs half of the heat rejected by engine $A$ (i.e.,$Q/2$) and rejects heat $Q_3$ to the sink at temperature $T_3$. When the work done in both cases is equal,the value of $T$ is:

When the temperature difference between the source and sink increases,the efficiency of the heat engine

The following figure shows a Carnot engine that works between temperatures $T_1=400 \text{ K}$ and $T_2=200 \text{ K}$ and drives a Carnot refrigerator that works between temperatures $T_3=350 \text{ K}$ and $T_4=250 \text{ K}$. The quantity $\frac{Q_3}{Q_1}$ will be

In a Carnot engine,when the temperatures are $T_2 = 0^{\circ} C$ and $T_1 = 200^{\circ} C$,its efficiency is $\eta_1$. When the temperatures are $T_1 = 0^{\circ} C$ and $T_2 = -200^{\circ} C$,its efficiency is $\eta_2$. Then the value of $\frac{\eta_1}{\eta_2}$ is: