A Carnot reversible engine converts $1/6$ of heat input into work. When the temperature of the sink is reduced by $62 \mathrm{K}$, the efficiency of Carnot's cycle becomes $1/3$. What is the sum of the temperature (in kelvin) of the source and the sink?

Helium gas goes through a cycle $\mathrm{ABCDA}$ (consisting of two isochoric and isobaric lines) as shown in figure. Efficiency of this cycle is nearly (Assume the gas to be close to ideal gas) 

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

When the absolute temperature of the source of a Carnot heat engine is increased by $25\%$ its efficiency increases by $80\%$. The new efficiency of the engine is

Consider a reversible engine of efficiency $\frac{1}{6}$. When the temperature of the sink is reduced by $62 °\mathrm{C}$, its efficiency gets doubled. The temperature of the source and sink respectively are 

An ideal gas is taken from state $1$ to state $2$ through optional path $A, B, C$ and $D$, as shown in the $PV$ diagram. Let $Q, W$ and $U$ represent the heat supplied, work done and change in internal energy of the gas, respectively. Then,