Efficiency of a Carnot engine is 26.81%. To increase its efficiency by 20%, its sink temperature will have to be____. The hot reservoir is maintained at 100 °C.

Options

A. increased by 20 K
B. decreased by 30 K
C. increased by 30 K
D. decreased by 20 K

Explanation

To solve the problem, we need to determine the new sink temperature (T2') of a Carnot engine when its efficiency is increased by 20% from the original efficiency of 26.81%. The hot reservoir temperature (T1) is given as 100 °C, which we will convert to Kelvin.

### Step-by-Step Solution:

1. **Convert the Hot Reservoir Temperature to Kelvin:**
\[ T1 = 100 °C = 100 + 273 = 373 \text{ K} \]

2. **Calculate the Original Efficiency:**
The original efficiency (η) is given as:
\[ η = 26.81\% = \frac{26.81}{100} = 0.2681 \]

3. **Calculate the New Efficiency:**
To increase the efficiency by 20%, we calculate the new efficiency (η'):
\[ η' = η + 0.20 \times η = 0.2681 + 0.20 \times 0.2681 = 0.2681 + 0.05362 = 0.32172 \]
Converting this back to percentage:
\[ η' = 32.172\% \]

4. **Use the Carnot Efficiency Formula:**
The efficiency of a Carnot engine is given by:
\[ η = 1 - \frac{T2}{T1} \]
Rearranging for T2:
\[ T2 = T1(1 - η) \]

5. **Calculate the Original Sink Temperature (T2):**
Using the original efficiency:
\[ T2 = 373 \times (1 - 0.2681) = 373 \times 0.7319 \approx 273 \text{ K} \]

6. **Calculate the New Sink Temperature (T2'):**
Now using the new efficiency:
\[ T2' = T1(1 - η') = 373 \times (1 - 0.32172) = 373 \times 0.67828 \approx 253 \text{ K} \]

7. **Determine the Change in Sink Temperature:**
The change in sink temperature (ΔT) is:
\[ ΔT = T2 - T2' = 273 - 253 = 20 \text{ K} \]

### Final Answer:
The sink temperature will have to decrease by 20 K to achieve the increased efficiency.