In a carnot engine, for `eta=1`, which of the following is true? (Symbols have their usual meaning)

Options

A. `T_(1)=T_(2)`
B. `T_(1)=0`
C. `T_(2)=0`
D. `eta` is independent of `T_(1) and T_(2)`.

Answer

To solve the problem, we need to analyze the efficiency of a Carnot engine and determine the conditions under which the efficiency (η) equals 1.

### Step-by-Step Solution:

1. **Understanding Carnot Engine Efficiency**:
The efficiency (η) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
where \(T_1\) is the temperature of the hot reservoir (source) and \(T_2\) is the temperature of the cold reservoir (sink).

2. **Setting Efficiency to 1**:
We are given that the efficiency is equal to 1:
\[ \eta = 1 \]
Substituting this into the efficiency formula gives:
\[ 1 = 1 - \frac{T_2}{T_1} \]

3. **Rearranging the Equation**:
Rearranging the equation leads to:
\[ 0 = -\frac{T_2}{T_1} \]
This implies:
\[ \frac{T_2}{T_1} = 0 \]

4. **Interpreting the Result**:
For the fraction \(\frac{T_2}{T_1}\) to equal 0, the numerator \(T_2\) must be equal to 0 (since \(T_1\) cannot be 0, as it would make the denominator undefined). Therefore, we conclude:
\[ T_2 = 0 \]

5. **Conclusion**:
Thus, for a Carnot engine to have an efficiency of 1, the temperature of the cold reservoir \(T_2\) must be 0.

### Final Answer:
The correct option is that \(T_2 = 0\).

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