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

  • A
    $\frac{T_1 T_2}{(1-\eta)(1-1.5 \eta)}$
  • B
    $\frac{0.5 T_2 \eta}{(1-1.5 \eta)(1-\eta)}$
  • C
    $\frac{T_1}{1-\eta}-\frac{T_2}{1-1.5 \eta}$
  • D
    $\frac{(1-\eta)(1-1.5 \eta)}{T_1 T_2}$

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The efficiency of an ideal heat engine working between the freezing point and boiling point of water is ........ $\%$

The $p-V$ diagram of a Carnot's engine is shown in the graph below. The engine uses $1$ mole of an ideal gas as the working substance. From the graph,the area enclosed by the $p-V$ diagram is equal to the net work done by the engine. Given that the heat supplied to the gas is $8000 \ J$,calculate the net work done by the engine. (Note: The efficiency of a Carnot engine is $\eta = 1 - \frac{T_2}{T_1} = \frac{W}{Q_1}$) (in $J$)

Explain heat engines and their working.

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The work done by a Carnot engine operating between $300 \,K$ and $400 \,K$ is $400 \,J$. The energy exhausted by the engine is (in $\,J$)

An engine operating between the boiling and freezing points of water will have:
$1.$ Efficiency more than $27 \%$
$2.$ Efficiency less than the efficiency of a Carnot engine operating between the same two temperatures.
$3.$ Efficiency equal to $27 \%$
$4.$ Efficiency less than $27 \%$

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