Question 9

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

Solution

We need to find the ratio $$\frac{\eta_1}{\eta_2}$$ for two Carnot engines.

Convert temperatures to Kelvin.

For engine 1: $$T_1 = 447 + 273 = 720 \text{ K}$$, $$T_2 = 147 + 273 = 420 \text{ K}$$

For engine 2: $$T_1 = 947 + 273 = 1220 \text{ K}$$, $$T_2 = 47 + 273 = 320 \text{ K}$$

Calculate the efficiencies.

$$\eta_1 = 1 - \frac{T_2}{T_1} = 1 - \frac{420}{720} = \frac{300}{720} = \frac{5}{12}$$ $$\eta_2 = 1 - \frac{T_2}{T_1} = 1 - \frac{320}{1220} = \frac{900}{1220} = \frac{45}{61}$$

Calculate the ratio.

$$\frac{\eta_1}{\eta_2} = \frac{5/12}{45/61} = \frac{5}{12} \times \frac{61}{45} = \frac{305}{540} = \frac{61}{108} \approx 0.5648 \approx 0.56$$

The correct answer is Option B: $$0.56$$.

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

Solution

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