In a water-fall the water falls from a height of 100 m . If the entire K.E. of water is converted into heat, the rise in temperature of water will be

Options

A. `0.23^(@)C`
B. `0.46^(@)C`
C. `2.3^(@)C`
D. `0.023^(@)C`

Answer

To solve the problem, we need to find the rise in temperature of water when it falls from a height of 100 m and all its kinetic energy is converted into heat.

### Step-by-Step Solution:

1. **Identify the potential energy (PE) of the water at the height**:
The potential energy of the water at height \( h \) is given by the formula:
\[ PE = mgh \]
where:
- \( m \) = mass of the water (in kg)
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( h \) = height (in meters)

For our case:
\[ h = 100 \, \text{m} \]
Therefore, the potential energy becomes:
\[ PE = mg \cdot 100 \]

2. **Convert potential energy to kinetic energy (KE)**:
When the water falls, all the potential energy is converted into kinetic energy. Thus:
\[ KE = PE = mgh \]

3. **Relate kinetic energy to heat energy**:
When the kinetic energy is converted into heat, it can be expressed as:
\[ KE = m \cdot S \cdot \Delta T \]
where:
- \( S \) = specific heat capacity of water (approximately \( 4200 \, \text{J/(kg} \cdot \text{°C)} \))
- \( \Delta T \) = rise in temperature (in °C)

4. **Set the equations equal to each other**:
From the above two equations, we have:
\[ mgh = m \cdot S \cdot \Delta T \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = S \cdot \Delta T \]

5. **Solve for \( \Delta T \)**:
Rearranging the equation gives:
\[ \Delta T = \frac{gh}{S} \]
Substituting the known values:
- \( g = 9.81 \, \text{m/s}^2 \)
- \( h = 100 \, \text{m} \)
- \( S = 4200 \, \text{J/(kg} \cdot \text{°C)} \)

Thus:
\[ \Delta T = \frac{9.81 \cdot 100}{4200} \]

6. **Calculate \( \Delta T \)**:
\[ \Delta T = \frac{981}{4200} \approx 0.233 \, \text{°C} \]

### Final Answer:
The rise in temperature of the water will be approximately \( 0.23 \, \text{°C} \).

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