For a Carnot engine T1 - T2 - When T2 is decreased by - T with T1 remaining same then efficiency is -1 and when T1 is increased by - T with T2 remaining same - efficiency is -2 which one of the following is the correct expression for -1 -2-

Η _1 η_2 = T_2T_1+Δ T T_2 Δ TT_1 =T_1T_2 (T_1+Δ T)(T_2 Δ T)T_1(T_1+Δ T) =T_1T_2[T_1T_2 +T_2Δ T T_1Δ T (Δ T)^2]T_1(T_1+Δ T) =(T_1 T_2)Δ T +(Δ T)^2T_1(T_1 +Δ T ...

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Byju's Answer

Standard XII

Chemistry

Electron Gain Enthalpy

For a Carnot ...

Question

For a Carnot engine $T_{1} > T_{2}$ . When $T_{2}$ is decreased by $Δ T$ with $T_{1}$ remaining same then efficiency is $η_{1}$ and when $T_{1}$ is increased by $Δ T$ with $T_{2}$ remaining same , efficiency is $η_{2}$ which one of the following is the correct expression for $η_{1}$ , $η_{2}$ ?

\frac{(T_{2} + T_{1}) Δ T + (Δ T)^{2}}{T_{1} (T_{1} + Δ T)}

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\frac{(T_{2} - T_{1}) Δ T + (Δ T)^{2}}{T_{1} (T_{1} + Δ T)}

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η_{1} = 1 - \frac{(T_{2} - Δ T)}{T_{1}}

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η_{2} = 1 - \frac{(T_{2})}{T_{1} + Δ T)}

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Solution

The correct option is A $\frac{(T_{2} + T_{1}) Δ T + (Δ T)^{2}}{T_{1} (T_{1} + Δ T)}$
$η_{1} - η_{2} = \frac{T_{2}}{T_{1} + Δ T} - \frac{T_{2} - Δ T}{T_{1}}$

$= \frac{T_{1} T_{2} - (T_{1} + Δ T) (T_{2} - Δ T)}{T_{1} (T_{1} + Δ T)}$

$= \frac{T_{1} T_{2} [T_{1} T_{2} + T_{2} Δ T - T_{1} Δ T - (Δ T)^{2}]}{T_{1} (T_{1} + Δ T)}$

$= \frac{(T_{1} - T_{2}) Δ T + (Δ T)^{2}}{T_{1} (T_{1} + Δ T)}$

$= \frac{(T_{2} - T_{1}) Δ T + (Δ T)^{2}}{T_{2} (T_{2} + Δ T)}$

$= \frac{(T_{1} + T_{2}) Δ T + (Δ T)^{2}}{T_{2} (T_{1} + Δ T)}$

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In a Carnot engine, when $T_{2} = 0^{\circ} C$ and $T_{1} = 200^{\circ} C$ , its efficiency is $η_{1}$ and when $T_{1} = 0^{\circ} C$ and $T_{2} = - 200^{\circ} C$ , Its efficiency is $η_{2}$ , then what is $\frac{η_{1}}{η_{2}}$