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

$A$ Carnot heat engine absorbs $600 \ J$ of heat from a source at a temperature of $127^{\circ} C$ and rejects $400 \ J$ of heat to a sink in each cycle. The temperature of the sink is (in $K$)

$A$ reversible Carnot heat engine converts $\frac{1}{4}$ of its input heat into work. When the temperature of the sink is reduced by $50 \ K$,its efficiency becomes $33 \frac{1}{3} \%$. The initial temperatures of the source and the sink respectively are

$A$ Carnot engine operating between temperatures $T_1$ and $T_2$ has an efficiency of $\frac{1}{6}$. When $T_2$ is lowered by $62\, K$,its efficiency increases to $\frac{1}{3}$. Then $T_1$ and $T_2$ are,respectively:

What is a Carnot cycle? Explain it by drawing a figure.

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 $\,K$)