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 can follow these instructions:

### Step 1: Understand the Efficiency of the Carnot Engine
The efficiency \( \eta \) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_C}{T_H} \]
where \( T_C \) is the temperature of the cold reservoir (low-temperature reservoir) and \( T_H \) is the temperature of the hot reservoir (high-temperature reservoir).

### Step 2: Convert Efficiency to Decimal
The efficiency is given as 25%, which can be expressed as:
\[ \eta = \frac{25}{100} = 0.25 \]

### Step 3: Rearrange the Efficiency Formula
Using the efficiency formula, we can rearrange it to find \( T_C \):
\[ 0.25 = 1 - \frac{T_C}{T_H} \]
This can be rewritten as:
\[ \frac{T_C}{T_H} = 1 - 0.25 = 0.75 \]
Thus, we can express \( T_C \) in terms of \( T_H \):
\[ T_C = 0.75 T_H \]

### Step 4: Use the Temperature Difference
We know that the temperature difference between the hot and cold reservoirs is given as:
\[ T_H - T_C = 80^\circ C \]
Substituting \( T_C \) from the previous step:
\[ T_H - 0.75 T_H = 80 \]
This simplifies to:
\[ 0.25 T_H = 80 \]

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

### Step 6: Calculate \( T_C \)
Now that we have \( T_H \), we can find \( T_C \):
\[ T_C = 0.75 T_H = 0.75 \times 320 = 240 \text{ degrees Celsius} \]

### Step 7: Convert to Celsius
Since the temperatures are typically expressed in Kelvin, we need to convert \( T_C \) from Kelvin to Celsius:
\[ T_C = 240 \text{ K} - 273 = -33^\circ C \]

### Final Answer
Thus, the temperature of the low-temperature reservoir \( T_C \) is:
\[ \boxed{-33^\circ C} \]