Calculate the difference in temperature of water at the top and the bottom of a water fall of 150 m height.

Answer

To calculate the difference in temperature of water at the top and the bottom of a waterfall of height 150 m, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to find the temperature difference (ΔT) between the water at the top and the bottom of a waterfall due to the potential energy converted into kinetic energy as the water falls.

2. **Identify the Relevant Formulas**:
- The work done by gravity on the water can be expressed as:
\[ W = Mgh \]
where:
- \(M\) = mass of water (in kg)
- \(g\) = acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\))
- \(h\) = height of the waterfall (150 m)

- The heat absorbed by the water can be expressed as:
\[ Q = Ms\Delta T \]
where:
- \(s\) = specific heat capacity of water (approximately \(4200 \, \text{J/kg°C}\))
- \(\Delta T\) = change in temperature (in °C)

3. **Set the Work Done Equal to Heat Absorbed**:
Since the work done by gravity is converted into heat absorbed by the water, we can set the two equations equal to each other:
\[ Mgh = Ms\Delta T \]

4. **Cancel Out Mass (M)**:
Since \(M\) appears on both sides of the equation, we can cancel it out:
\[ gh = s\Delta T \]

5. **Rearrange to Solve for ΔT**:
Rearranging the equation gives:
\[ \Delta T = \frac{gh}{s} \]

6. **Substitute Known Values**:
- \(g = 9.81 \, \text{m/s}^2\)
- \(h = 150 \, \text{m}\)
- \(s = 4200 \, \text{J/kg°C}\)

Substituting these values into the equation:
\[ \Delta T = \frac{(9.81 \, \text{m/s}^2)(150 \, \text{m})}{4200 \, \text{J/kg°C}} \]

7. **Calculate ΔT**:
\[ \Delta T = \frac{1471.5 \, \text{m}^2/\text{s}^2}{4200 \, \text{J/kg°C}} \approx 0.35 \, \text{°C} \]

### Final Answer:
The difference in temperature of water at the top and the bottom of the waterfall is approximately **0.35 °C**.

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