Water falls on the ground from 42 m height. It half of loss is potential energy converts into heat, then increase in temperature of water is

Options

A. `205.8^(@)` C
B. `3.49^(@)` C
C. `0.049^(@)` C
D. `0.098^(@)` C

Explanation

To solve the problem, we need to calculate the increase in temperature of water when it falls from a height of 42 meters, given that half of the loss in potential energy is converted into heat. Here's the step-by-step solution:

### Step 1: Calculate the potential energy lost by the water
The potential energy (PE) lost by the water when it falls from a height \( h \) is given by 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 (42 m)

### Step 2: Calculate the heat energy converted from potential energy
According to the problem, half of the potential energy is converted into heat energy \( Q \):
\[ Q = \frac{1}{2} mgh \]

### Step 3: Relate heat energy to temperature change
The heat energy absorbed by the water can also be expressed in terms of the mass of the water, its specific heat capacity \( c \), and the change in temperature \( \Delta T \):
\[ Q = mc\Delta T \]
where:
- \( c \) = specific heat capacity of water (approximately \( 4.2 \times 10^3 \, \text{J/kg°C} \))

### Step 4: Set the two expressions for heat energy equal to each other
From the previous steps, we have:
\[ \frac{1}{2} mgh = mc\Delta T \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} gh = c\Delta T \]

### Step 5: Solve for the change in temperature \( \Delta T \)
Rearranging the equation gives us:
\[ \Delta T = \frac{1}{2} \frac{gh}{c} \]

### Step 6: Substitute the known values
Now, substituting \( g = 9.8 \, \text{m/s}^2 \), \( h = 42 \, \text{m} \), and \( c = 4.2 \times 10^3 \, \text{J/kg°C} \):
\[ \Delta T = \frac{1}{2} \cdot \frac{9.8 \cdot 42}{4.2 \times 10^3} \]

### Step 7: Calculate the numerical value
Calculating the numerator:
\[ 9.8 \cdot 42 = 411.6 \]
Now substituting back:
\[ \Delta T = \frac{1}{2} \cdot \frac{411.6}{4200} \]
Calculating:
\[ \Delta T = \frac{411.6}{8400} \approx 0.049 \, °C \]

### Final Result
The increase in temperature of the water is approximately:
\[ \Delta T \approx 0.049 \, °C \]