Question

A Carnot engine whose heat sinks at 27°C, has an efficiency of 25%. By how many degrees should the temperature of the source be changed to increase the efficiency by 100% of the original efficiency?

• A. Increases by 18°C
• B. Increases by 200°C
• C. Increases by 120°C
• D. Increases by 73°C

Answer 1

The Correct Option is B

To solve this problem, we need to understand the working of a Carnot engine and the relationship between efficiency, source temperature, and sink temperature. The efficiency \eta of a Carnot engine is given by the formula:

\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}

where T_{\text{sink}} and T_{\text{source}} are the absolute temperatures of the sink and source, respectively, measured in Kelvin.

1. Convert the sink temperature from Celsius to Kelvin: T_{\text{sink}} = 27 + 273 = 300 \, K.
2. The initial efficiency is given as 25%, or 0.25 in decimal form. Using the efficiency formula:
   0.25 = 1 - \frac{300}{T_{\text{source}}}
   Solving for T_{\text{source}}, we get:
   1 - 0.25 = \frac{300}{T_{\text{source}}}
   0.75 = \frac{300}{T_{\text{source}}} T_{\text{source}} = \frac{300}{0.75} = 400 \, K
3. We need to increase the efficiency by 100% of the original efficiency. Since the original efficiency is 25%, a 100% increase would make the new efficiency 50%, or 0.50 in decimal form.
4. Apply this new efficiency to the Carnot efficiency formula:
   0.50 = 1 - \frac{300}{T_{\text{new source}}}
   Solving for T_{\text{new source}}, we have:
   1 - 0.50 = \frac{300}{T_{\text{new source}}}
   0.50 = \frac{300}{T_{\text{new source}}} T_{\text{new source}} = \frac{300}{0.50} = 600 \, K
5. The change in source temperature required is the difference between the new source temperature and the original source temperature: T_{\text{change}} = 600 \, K - 400 \, K = 200 \, K.

Therefore, the temperature of the source should be increased by 200°C. Hence, the correct answer is:

Increases by 200°C