What is the source temperature of the Carnot engine required to get $70 \%$ efficiency (in $^{\circ}C$)? Given sink temperature $= 27^{\circ}C$.

$A$ car engine has a power of $20 kW$. The car makes a round trip of $1$ hour. If the thermal efficiency of the engine is $40 \%$ and the ambient temperature is $300 K$, the energy generated by fuel combustion is: (in $kJ$)

For a Carnot engine,the efficiency is given by $W/Q_1 = 1/6$. If the temperature of the sink is reduced by $62^{\circ}C$,the efficiency doubles. The initial temperatures of the sink and the source are,respectively:

The efficiency of a reversible engine is . . . . . . the efficiency of an irreversible engine.

$A$ 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:

$A$ 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 $K$)