Let $\eta_1$ is the efficiency of an engine at $T_1=447^{\circ}\mathrm{C}$ and $T_2=147^{\circ}\mathrm{C}$ while $\eta_2$ is the efficiency at $T_1=947^{\circ}\mathrm{C}$ and $T_2=47^{\circ}\mathrm{C}$. The ratio $\frac{\eta_1}{\eta_2}$ will be:

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Heat & Thermodynamics

Let

η_{1}

is the efficiency of an engine at

T_{1} = 44 7^{\circ} C

and

T_{2} = 14 7^{\circ} C

while

η_{2}

is the efficiency at

T_{1} = 94 7^{\circ} C

and

T_{2} = 4 7^{\circ} C

. The ratio

\frac{η _{1}}{η _{2}}

will be:

(A)0.41
(B)0.56✓ Correct
(C)0.73
(D)0.70

Answer: (B)

Solution

For a Carnot engine, efficiency

η = 1 - \frac{T _{L}}{T _{H}}

(temperatures in Kelvin).

For the first engine:

η_{1} = 1 - \frac{147 + 273}{447 + 273} = 1 - \frac{420}{720} = \frac{300}{720}

For the second engine:

η_{2} = 1 - \frac{47 + 273}{947 + 273} = 1 - \frac{320}{1220} = \frac{900}{1220}

The ratio:

\frac{η _{1}}{η _{2}} = \frac{300}{720} \times \frac{1220}{900} = \frac{300 \times 1220}{720 \times 900} = \frac{122}{216} \approx 0.56

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option

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