The efficiency of a Carnot cycle is $\frac{1}{6}$. By lowering the temperature of the sink by $65 \ K$,it increases to $\frac{1}{3}$. The initial and final temperatures of the sink are:

$A$ Carnot engine operating between temperatures $600 \,K$ and $300 \,K$ absorbs $800 \,J$ of heat from the source. The mechanical work done per cycle is (in $\,J$)

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:

Determine the efficiency of a Carnot cycle if,in the adiabatic expansion,the volume becomes $3$ times its initial value and $\gamma = 1.5$.

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

Two heat engines $X$ and $Y$ of same efficiency are connected in series in such a way that the sink of $X$ works as the source of $Y$. $X$ receives heat at $900 \,K$ and rejects some heat to its sink at $T \,K$, and in turn, $Y$ rejects heat to its sink at $400 \,K$. Then the temperature $T$ is: (in $\,K$)