At what temperature will the speed of sound be double of its value at `0^@C ` ?

Explanation

To solve the problem of finding the temperature at which the speed of sound is double its value at 0°C, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between speed of sound and temperature**:
The speed of sound (V) in air is proportional to the square root of the absolute temperature (T). This can be expressed as:
\[ V \propto \sqrt{T} \]

2. **Set up the initial conditions**:
At 0°C (which is 273 K), let the speed of sound be \( V_1 \). Thus, we have:
\[ T_1 = 273 \, \text{K} \]
\[ V_1 = V \]

3. **Determine the new speed of sound**:
We want to find the temperature \( T_2 \) at which the speed of sound \( V_2 \) is double that of \( V_1 \):
\[ V_2 = 2V \]

4. **Use the proportionality relationship**:
Since \( V \propto \sqrt{T} \), we can write:
\[ \frac{V_2}{V_1} = \frac{\sqrt{T_2}}{\sqrt{T_1}} \]
Substituting \( V_2 = 2V \) and \( V_1 = V \):
\[ \frac{2V}{V} = \frac{\sqrt{T_2}}{\sqrt{T_1}} \]
This simplifies to:
\[ 2 = \frac{\sqrt{T_2}}{\sqrt{T_1}} \]

5. **Square both sides**:
Squaring both sides gives:
\[ 4 = \frac{T_2}{T_1} \]

6. **Solve for \( T_2 \)**:
Rearranging the equation:
\[ T_2 = 4 \cdot T_1 \]
Substituting \( T_1 = 273 \, \text{K} \):
\[ T_2 = 4 \cdot 273 = 1092 \, \text{K} \]

7. **Convert to Celsius**:
To convert Kelvin to Celsius:
\[ T_2 (\text{°C}) = T_2 (\text{K}) - 273 \]
Thus,
\[ T_2 = 1092 - 273 = 819 \, \text{°C} \]

### Final Answer:
The temperature at which the speed of sound is double its value at 0°C is **819°C**.

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