To increase the efficiency of a carnot engine, will you prefer to (i) increase the temp. of source by`10K` or (ii) decrease the temp. of sink by `10K`?

Answer

To solve the problem of whether to increase the temperature of the source by 10K or decrease the temperature of the sink by 10K to increase the efficiency of a Carnot engine, we can follow these steps:

### Step 1: Understand the Efficiency Formula
The efficiency (η) of a Carnot engine is given by the formula:
\[ η = 1 - \frac{T_2}{T_1} \]
where:
- \(T_1\) is the temperature of the source (hot reservoir),
- \(T_2\) is the temperature of the sink (cold reservoir).

### Step 2: Calculate Efficiency After Increasing Source Temperature
If we increase the temperature of the source by 10K, the new temperature of the source becomes:
\[ T_1' = T_1 + 10 \]
The new efficiency (η₁) will then be:
\[ η_1 = 1 - \frac{T_2}{T_1 + 10} \]

### Step 3: Calculate Efficiency After Decreasing Sink Temperature
If we decrease the temperature of the sink by 10K, the new temperature of the sink becomes:
\[ T_2' = T_2 - 10 \]
The new efficiency (η₂) will then be:
\[ η_2 = 1 - \frac{T_2 - 10}{T_1} \]

### Step 4: Compare the Two Efficiencies
Now we need to compare η₁ and η₂:
1. Substitute η₁ and η₂:
\[ η_1 = 1 - \frac{T_2}{T_1 + 10} \]
\[ η_2 = 1 - \frac{T_2 - 10}{T_1} \]

2. To find out which efficiency is greater, we can calculate the difference:
\[ η_2 - η_1 = \left(1 - \frac{T_2 - 10}{T_1}\right) - \left(1 - \frac{T_2}{T_1 + 10}\right) \]
Simplifying this gives:
\[ η_2 - η_1 = \frac{T_2}{T_1 + 10} - \frac{T_2 - 10}{T_1} \]

### Step 5: Simplify the Expression
To simplify:
1. Combine the fractions:
\[ η_2 - η_1 = \frac{T_2 T_1 - (T_2 - 10)(T_1 + 10)}{T_1(T_1 + 10)} \]

2. Expand and simplify the numerator:
\[ = \frac{T_2 T_1 - (T_2 T_1 + 10T_2 - 10T_1 - 100)}{T_1(T_1 + 10)} \]
\[ = \frac{10T_1 - 10T_2 + 100}{T_1(T_1 + 10)} \]

### Step 6: Analyze the Result
The expression \(10(T_1 - T_2 + 10)\) indicates that if \(T_1 > T_2\), then \(η_2 - η_1 > 0\). This means that decreasing the temperature of the sink by 10K results in a greater increase in efficiency than increasing the temperature of the source by 10K.

### Conclusion
Thus, to increase the efficiency of a Carnot engine, it is preferable to **decrease the temperature of the sink by 10K**.

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