Water falls from a height 500 m. what is the rise in temperature of water at bottom if whole energy remains in the water?

Options

A. `0.96^(@)C`
B. `1.02^(@)C`
C. `1.16^(@)C`
D. `0.23^(@)C`

Answer

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

### Step 1: Understand the Energy Conversion
When water falls from a height, its potential energy is converted into heat energy. The potential energy (PE) at height \( h \) is given by the formula:
\[ PE = mgh \]
where:
- \( m \) = mass of the water (in kg)
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( h \) = height (in meters)

### Step 2: Write the Heat Energy Equation
The heat energy (\( Q \)) gained by the water when it reaches the bottom can be expressed as:
\[ Q = mc\Delta T \]
where:
- \( c \) = specific heat capacity of water (approximately \( 4.18 \, \text{kJ/kg°C} \) or \( 4180 \, \text{J/kg°C} \))
- \( \Delta T \) = rise in temperature (in °C)

### Step 3: Set the Potential Energy Equal to Heat Energy
Since all potential energy is converted into heat energy, we can set the two equations equal to each other:
\[ mgh = mc\Delta T \]

### Step 4: Cancel Out the Mass
Assuming the mass \( m \) is not zero, we can cancel \( m \) from both sides of the equation:
\[ gh = c\Delta T \]

### Step 5: Solve for \( \Delta T \)
Rearranging the equation to solve for \( \Delta T \):
\[ \Delta T = \frac{gh}{c} \]

### Step 6: Substitute Known Values
Now we can substitute the known values into the equation:
- \( g = 9.81 \, \text{m/s}^2 \)
- \( h = 500 \, \text{m} \)
- \( c = 4180 \, \text{J/kg°C} \)

Substituting these values:
\[ \Delta T = \frac{(9.81 \, \text{m/s}^2)(500 \, \text{m})}{4180 \, \text{J/kg°C}} \]

### Step 7: Calculate \( \Delta T \)
Calculating the numerator:
\[ 9.81 \times 500 = 4905 \, \text{J/kg} \]
Now substituting into the equation:
\[ \Delta T = \frac{4905}{4180} \approx 1.17 \, \text{°C} \]

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

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