Calculate the rise in temperature of water, which falls from a height of 100m. Assume that `80%` of the energy due to fall is converted into heat and is retaied by the water. `J = 4.2 xx 10^(7) erg cal^(-1)`.

Answer

To solve the problem of calculating the rise in temperature of water that falls from a height of 100 meters, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Height (h) = 100 m
- Percentage of energy converted to heat = 80% = 0.8
- Gravitational acceleration (g) = 9.8 m/s²
- Conversion factor: 1 calorie = 4.2 joules

2. **Calculate the Potential Energy (PE):**
The potential energy (PE) of the water at height h is given by the formula:
\[ PE = mgh \]
where:
- m = mass of water (in kg)
- g = acceleration due to gravity (9.8 m/s²)
- h = height (100 m)

3. **Calculate the Heat Energy Retained:**
Since 80% of the potential energy is converted into heat, the heat energy (Q) retained by the water is:
\[ Q = 0.8 \times PE = 0.8 \times mgh \]

4. **Relate Heat Energy to Temperature Change:**
The heat energy retained can also be expressed in terms of the mass of water and its specific heat capacity (c):
\[ Q = mc\Delta T \]
where:
- c = specific heat capacity of water = 1 cal/g = 4.2 × 10³ J/kg (after converting to SI units)
- \(\Delta T\) = change in temperature (in °C or K)

5. **Set the Equations Equal:**
Equating the two expressions for Q, we have:
\[ 0.8 \times mgh = mc\Delta T \]

6. **Cancel Mass (m):**
Since mass (m) appears on both sides, we can cancel it (assuming m ≠ 0):
\[ 0.8gh = c\Delta T \]

7. **Solve for \(\Delta T\):**
Rearranging the equation to solve for the change in temperature (\(\Delta T\)):
\[ \Delta T = \frac{0.8gh}{c} \]

8. **Substitute the Values:**
Now we substitute the known values into the equation:
\[ \Delta T = \frac{0.8 \times 9.8 \, \text{m/s}^2 \times 100 \, \text{m}}{4.2 \times 10^3 \, \text{J/kg}} \]

9. **Calculate \(\Delta T\):**
Performing the calculation:
\[ \Delta T = \frac{0.8 \times 9.8 \times 100}{4.2 \times 10^3} \]
\[ \Delta T = \frac{784}{4200} \approx 0.18667 \, \text{°C} \]

10. **Round the Result:**
Rounding the result gives:
\[ \Delta T \approx 0.187 \, \text{°C} \]

### Final Answer:
The rise in temperature of the water is approximately **0.187 °C**.

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