The efficiency of an ideal Carnot engine working between temperatures $T_1$ and $T_2$ is $1/3$. If the temperature of the sink is reduced by $40 \%$,then its efficiency will be: (in $\%$)

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 $\,K$)

The efficiency of a Carnot engine is $\eta$ when its hot and cold reservoirs are maintained at temperatures $T_1$ and $T_2$, respectively. To increase the efficiency to $1.5 \eta$, the increase in temperature $(\Delta T)$ of the hot reservoir, while keeping the cold reservoir constant at $T_2$, is

The temperature of food material in a refrigerator is $4^{\circ} C$ and the temperature of the environment is $15^{\circ} C$. If a Carnot cycle is used in its working gas,find its Carnot efficiency.

An ideal heat engine operates in a Carnot cycle between $127^{\circ} C$ and $27^{\circ} C$. It absorbs $5 \times 10^4 \text{ cal}$ of heat at the higher temperature. The amount of heat converted to work is:

$A$ Carnot engine takes $3 \times 10^6 \, \text{cal}$ of heat from a reservoir at $627^{\circ}C$ and gives it to a sink at $27^{\circ}C$. The work done by the engine is: