Water in a fall falls through h=1.0 km considering this as a thermodynamical process, calculate the difference in temperature of water at the top and bottom of the fall. Specific heat of water `=1000"cal"//kg^(@)C` and J=4.2J/cal.

Answer

To solve the problem of calculating the temperature difference of water falling from a height of 1.0 km, we will use the principles of energy conservation, specifically the conversion of potential energy to thermal energy.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Height of the fall, \( h = 1.0 \, \text{km} = 1000 \, \text{m} \)
- Specific heat of water, \( c = 1000 \, \text{kcal/kg} \cdot \degree C \)
- Conversion factor, \( 1 \, \text{kcal} = 4.2 \, \text{J} \)
- Acceleration due to gravity, \( g \approx 10 \, \text{m/s}^2 \)

2. **Calculate the Potential Energy (PE):**
The potential energy lost by the water as it falls is given by the formula:
\[ PE = mgh \]
where \( m \) is the mass of the water, \( g \) is the acceleration due to gravity, and \( h \) is the height.

3. **Convert Potential Energy to Heat Energy (Q):**
The heat energy gained by the water is given by:
\[ Q = mc\Delta T \]
where \( \Delta T = T_2 - T_1 \) is the change in temperature.

4. **Set the Potential Energy Equal to Heat Energy:**
Since the potential energy lost is converted into heat energy gained, we can equate the two:
\[ mgh = mc\Delta T \]

5. **Cancel the Mass (m):**
Since mass \( m \) appears on both sides of the equation, we can cancel it out:
\[ gh = c\Delta T \]

6. **Rearrange to Solve for \( \Delta T \):**
\[ \Delta T = \frac{gh}{c} \]

7. **Substitute the Values:**
Convert the specific heat from kcal to joules:
\[ c = 1000 \, \text{kcal/kg} \cdot \degree C \times 4.2 \, \text{J/kcal} = 4200 \, \text{J/kg} \cdot \degree C \]
Now substitute the values into the equation:
\[ \Delta T = \frac{(10 \, \text{m/s}^2)(1000 \, \text{m})}{4200 \, \text{J/kg} \cdot \degree C} \]

8. **Calculate \( \Delta T \):**
\[ \Delta T = \frac{10000 \, \text{m}^2/\text{s}^2}{4200 \, \text{J/kg} \cdot \degree C} \approx 2.38 \, \degree C \]

9. **Final Result:**
The difference in temperature of the water at the top and bottom of the fall is approximately:
\[ \Delta T \approx 2.38 \, \degree C \]