The temperature of the source of a Carnot engine operating with an efficiency of $70\%$ is $1000 \ K$. The temperature of its sink is ...... $K$.

$A$ Carnot engine of efficiency $40 \%$ takes heat from a source maintained at a temperature of $500 \ K$. It is desired to have an engine of efficiency $60 \%$. Then,the source temperature for the same sink temperature must be (in $K$)

An engine has an efficiency of $1/6$. When the temperature of the sink is reduced by $62^{\circ}C$,its efficiency is doubled. The temperature of the source is ....... $^{\circ}C$.

In practice,all heat engines have an efficiency less than that of a Carnot engine because

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:

$A$ Carnot engine absorbs $1000\,J$ of heat energy from a reservoir at $127\,^oC$ and rejects $600\,J$ of heat energy during each cycle. The efficiency of the engine and the temperature of the sink will be: