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(B) The efficiency $(\eta)$ of a Carnot engine is given by the formula:$\eta = 1 - \frac{T_{L}}{T_{H}}$where $T_{L}$ is the temperature of the cold re

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(B) The efficiency $(\eta)$ of a Carnot engine is given by the formula:$\eta = 1 - \frac{T_{L}}{T_{H}}$where $T_{L}$ is the temperature of the cold reservoir and $T_{H}$ is the temperature of the hot reservoir.For a fixed cold reservoir temperature $T_{L}$,as the hot reservoir temperature $T_{H}$ increases,the ratio $\frac{T_{L}}{T_{H}}$ decreases.Consequently,the efficiency $\eta = 1 - \frac{T_{L}}{T_{H}}$ increases as $T_{H}$ increases.When $T_{H} = T_{L}$,the efficiency $\eta = 1 - \frac{T_{L}}{T_{L}} = 0$.As $T_{H} 	o \infty$,the efficiency $\eta 	o 1$.The graph of $\eta$ versus $T_{H}$ starts from $0$ at $T_{H} = T_{L}$ and increases with a decreasing slope (concave down) towards the value $1$.This behavior is correctly represented by the graph in option $(b)$.

$A$ number of Carnot engines are operated at identical cold reservoir temperatures $(T_{L})$. However,their hot reservoir temperatures are kept different. $A$ graph of the efficiency of the engines versus hot reservoir temperature $(T_{H})$ is plotted. The correct graphical representation is

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An engine operating between the boiling and freezing points of water will have:
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