The efficiency of a Carnot engine depends on what,and of what is it independent?

$A$ diatomic ideal gas is used as the working substance in a Carnot engine. During the adiabatic expansion process,if the volume of the gas increases from $V$ to $32V$,what is the efficiency of the engine?

Let $\eta_{1}$ be the efficiency of a Carnot engine at $T_{H}=447^{\circ}C$ and $T_{L}=147^{\circ}C$,while $\eta_{2}$ is the efficiency at $T_{H}=947^{\circ}C$ and $T_{L}=47^{\circ}C$. The ratio $\frac{\eta_{1}}{\eta_{2}}$ will be:

Consider a two-stage Carnot engine. In the first stage,heat $Q_1$ is absorbed at temperature $T$ and heat $Q_2$ is expelled at temperature $\alpha T$ (where $\alpha < 1$). In the second stage,heat $Q_2$ is absorbed at temperature $\alpha T$ and heat $Q_3$ is expelled at temperature $\beta T$ (where $\beta < \alpha$). The efficiency of the Carnot engine will be

Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures $T_{1}$ and $T_{2}$. The temperature of the hot reservoir of the first engine is $T_{1}$ and the temperature of the cold reservoir of the second engine is $T_{2}$. $T$ is the temperature of the sink of the first engine,which is also the source for the second engine. How is $T$ related to $T_{1}$ and $T_{2}$,if both engines perform equal amounts of work?

$A$ Carnot engine has an efficiency of $50 \%$. If the temperature of the sink is reduced by $40 \, K$,its efficiency increases by $30 \%$. The temperature of the source will be $.... \, K$.