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:

The efficiency of a Carnot engine is $50\%$ and the temperature of the sink is $500 \ K$. If the temperature of the source is kept constant and its efficiency is raised to $60\%$,then the required temperature of the sink will be: (in $K$)

$A$ Carnot engine $(E)$ is working between two temperatures $473 \ K$ and $273 \ K$. In a new system,two engines are used: engine $E_1$ works between $473 \ K$ and $373 \ K$,and engine $E_2$ works between $373 \ K$ and $273 \ K$. If $\eta_{12}$,$\eta_1$,and $\eta_2$ are the efficiencies of the engines $E$,$E_1$,and $E_2$ respectively,then:

$A$ Carnot engine has an efficiency of $\frac{1}{6}$. When the temperature of the sink is lowered by $57 \ K$,its efficiency becomes $\frac{1}{3}$. The temperature of the source is: (in $K$)

Suppose that two heat engines are connected in series,such that the heat released by the first engine is used as the heat absorbed by the second engine,as shown in the figure. The efficiencies of the engines are $\epsilon_1$ and $\epsilon_2$,respectively. The net efficiency of the combination is given by:

Efficiency of a heat engine whose sink is at temperature of $300 \,K$ is $40 \%$. To increase the efficiency to $60 \%$, keeping the sink temperature constant, the source temperature must be increased by (in $\,K$)