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Two Carnot engines A and B are operated in series. Engine A receives heat from a reservoir at 600 K and rejects heat to a reservoir at temperature T. Engine B receives heat rejected by engine A and in turn rejects it to a reservoir at 100 K. If the efficiencies of the two engines A and B are represented by ηA and ηB, respectively,then what is the value of $\frac{\eta_{B} }{\eta _{A}}$ ?

Option 1)
$\frac{12}{7}$

Option 2)
$\frac{7}{12}$

Option 3)
$\frac{12}{5}$

Option 4)
$\frac{5}{12}$

Answers (2)

best_answer

As we learnt

Efficiency of a carnot cycle -

$\eta =\frac{W}{Q_{1}-Q_{2}}=1-\frac{T_{2}}{T_{1}}$

$T_{1}\, and\, T_{2}$ are in kelvin

- wherein

$T_{1}=$ Source temperature

$T_{2}=$ Sink Temperature

$\left ( T_{1} > T_{2}\right )$

$\eta _{A}=1-\frac{T}{600}=\frac{w}{Q_{1}}\rightarrow (1)$
$\eta _{B}=1-\frac{100}{T}=\frac{w}{Q_{2}}=\frac{w}{Q_{1-w}}$

$\eta _{B}=1-\frac{100}{T}=\frac{1}{\frac{Q_{1}}{w}-1}$

from eqn1:

$\eta _{B}=1-\frac{100}{T}=\frac{1}{\frac{1}{1-\frac{T}{600}}-1}=\frac{1-\frac{T}{600}}{\frac{T}{600}}$

$\eta _{B}=\frac{600}{T}-1= 1-\frac{100}{T}\: \: or\: \: \frac{700}{T}=2\: \: or\: \: T=350K$

$\frac{\eta _{B}}{\eta _{A}}= \frac{1-\frac{100}{T}}{1-\frac{T}{600}}=\frac{12}{7}$

Option 1)

$\frac{12}{7}$

Option 2)

$\frac{7}{12}$

Option 3)

$\frac{12}{5}$

Option 4)

$\frac{5}{12}$

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Aadil

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