A water fall is 84 m high. Assuming that half the kinetic energy of falling water get converted to heat, the rise in temperature of water is

Options

A. `0.098^(@)C`
B. `0.98^(@)C`
C. `98^(@)C`
D. `0.0098^(@)C`

Answer

To solve the problem, we need to determine the rise in temperature of water falling from a height of 84 meters, given that half of its kinetic energy is converted to heat energy.

### Step-by-step Solution:

1. **Calculate the Potential Energy (PE) at the height of 84 m:**
The potential energy (PE) of the water at height \( h \) can be calculated using the formula:
\[ PE = mgh \]
where:
- \( m \) = mass of the water (in kg)
- \( g \) = acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \))
- \( h \) = height (in meters)

Substituting the values:
\[ PE = mg \cdot 84 \]

2. **Calculate the Kinetic Energy (KE) just before hitting the ground:**
The kinetic energy (KE) just before the water hits the ground is equal to the potential energy at the height (assuming no energy loss):
\[ KE = PE = mgh = mg \cdot 84 \]

3. **Determine the Heat Energy (Q) converted from KE:**
According to the problem, half of the kinetic energy is converted to heat energy:
\[ Q = \frac{1}{2} KE = \frac{1}{2} (mgh) = \frac{1}{2} (mg \cdot 84) \]

4. **Relate Heat Energy to Temperature Rise:**
The heat energy can also be expressed in terms of mass, specific heat capacity, and temperature change:
\[ Q = mc\Delta T \]
where:
- \( c \) = specific heat capacity of water (approximately \( 4200 \, \text{J/kg°C} \))
- \( \Delta T \) = rise in temperature (in °C)

5. **Set the two expressions for Q equal to each other:**
\[ \frac{1}{2} (mg \cdot 84) = mc\Delta T \]

6. **Cancel mass (m) from both sides:**
\[ \frac{1}{2} g \cdot 84 = c \Delta T \]

7. **Solve for ΔT:**
Rearranging gives:
\[ \Delta T = \frac{\frac{1}{2} g \cdot 84}{c} \]
Substituting \( g = 9.8 \, \text{m/s}^2 \) and \( c = 4200 \, \text{J/kg°C} \):
\[ \Delta T = \frac{\frac{1}{2} \cdot 9.8 \cdot 84}{4200} \]

8. **Calculate ΔT:**
\[ \Delta T = \frac{4.9 \cdot 84}{4200} = \frac{411.6}{4200} \approx 0.098 \, °C \]

### Final Answer:
The rise in temperature of the water is approximately \( 0.098 \, °C \).

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