Water falls from a height `500m`, what is the rise in temperature of water at bottom if whole energy remains in the water ? `(J=4.2)`

Options

A. `0.96^(@)C`
B. `1.02^(@)C`
C. `1.16^(@)C`
D. `0.3^(@)C`

Explanation

To solve the problem of finding the rise in temperature of water when it falls from a height of 500 meters, we can follow these steps:

### Step 1: Understand the Energy Conversion
When water falls from a height, its potential energy is converted into kinetic energy and then into heat energy when it reaches the bottom. We will use the potential energy formula to find the energy converted into heat.

### Step 2: Write the Formula for Potential Energy
The potential energy (PE) of the water at height \( h \) is given by:
\[ PE = mgh \]
where:
- \( m \) = mass of water (in kg)
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( h \) = height (in meters)

For our case, \( h = 500 \, \text{m} \).

### Step 3: Calculate the Potential Energy
Substituting the values into the potential energy formula:
\[ PE = m \cdot 9.81 \cdot 500 \]
\[ PE = 4905m \, \text{Joules} \]

### Step 4: Relate Heat Energy to Temperature Change
The heat energy gained by the water can be expressed as:
\[ Q = mc\Delta T \]
where:
- \( Q \) = heat energy (in Joules)
- \( c \) = specific heat capacity of water (approximately \( 4200 \, \text{J/kg°C} \))
- \( \Delta T \) = rise in temperature (in °C)

### Step 5: Set the Potential Energy Equal to Heat Energy
Since all potential energy is converted into heat energy:
\[ 4905m = mc\Delta T \]

### Step 6: Cancel the Mass \( m \)
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ 4905 = c\Delta T \]

### Step 7: Substitute the Value of Specific Heat Capacity
Substituting \( c = 4200 \, \text{J/kg°C} \):
\[ 4905 = 4200 \Delta T \]

### Step 8: Solve for \( \Delta T \)
Now, we can solve for \( \Delta T \):
\[ \Delta T = \frac{4905}{4200} \]
\[ \Delta T \approx 1.1667 \, \text{°C} \]

### Step 9: Round the Result
Rounding this to two decimal places gives:
\[ \Delta T \approx 1.17 \, \text{°C} \]

### Final Answer
The rise in temperature of the water at the bottom is approximately \( 1.17 \, \text{°C} \).

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