A Carnot engine has efficiency 25%. It operates between reservoirs of constant temperatuers with temperature difference of `80^@C`. What is the temperature of the low -temperature reservoir ?

Options

A. `-25^@C`
B. `25^@C`
C. `-33^@C`
D. `33^@C`

Answer

To solve the problem step by step, we will use the relationship between the efficiency of a Carnot engine and the temperatures of the hot and cold reservoirs.

### Step 1: Understand the given information
- Efficiency (η) of the Carnot engine = 25% = 0.25
- Temperature difference between the reservoirs = 80°C

### Step 2: Use the formula for efficiency of a Carnot engine
The efficiency 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)
- \(T_2\) is the temperature of the cold reservoir (sink)

### Step 3: Rearrange the efficiency formula
From the efficiency formula, we can express \(T_2\) in terms of \(T_1\):
\[ \frac{T_2}{T_1} = 1 - \eta \]
Substituting the value of efficiency:
\[ \frac{T_2}{T_1} = 1 - 0.25 = 0.75 \]
Thus, we have:
\[ T_2 = 0.75 T_1 \]

### Step 4: Use the temperature difference
The problem states that the temperature difference between the two reservoirs is 80°C:
\[ T_1 - T_2 = 80 \]
Substituting \(T_2\) from the previous step:
\[ T_1 - 0.75 T_1 = 80 \]
This simplifies to:
\[ 0.25 T_1 = 80 \]

### Step 5: Solve for \(T_1\)
Now, we can solve for \(T_1\):
\[ T_1 = \frac{80}{0.25} = 320 \text{ K} \]

### Step 6: Find \(T_2\)
Now that we have \(T_1\), we can find \(T_2\):
\[ T_2 = 0.75 T_1 = 0.75 \times 320 = 240 \text{ K} \]

### Step 7: Convert \(T_2\) to Celsius
To convert Kelvin to Celsius, we use the formula:
\[ T(°C) = T(K) - 273 \]
Thus,
\[ T_2(°C) = 240 - 273 = -33°C \]

### Final Answer
The temperature of the low-temperature reservoir \(T_2\) is \(-33°C\).

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