A `50 kg` man is running at a speed of `18 kmH^(-1)` If all the kinetic energy of the man be uses to increase the temperature of water from `30^(@)C`how much water can be beated with this energy?

Explanation

To solve the problem step by step, we will calculate the kinetic energy of the man and then determine how much water can be heated using that energy.

### Step 1: Convert the speed from km/h to m/s
The speed of the man is given as \( 18 \, \text{km/h} \). To convert this to meters per second (m/s), we use the conversion factor \( 1 \, \text{km/h} = \frac{5}{18} \, \text{m/s} \).

\[ v = 18 \, \text{km/h} \times \frac{5}{18} = 5 \, \text{m/s} \]

### Step 2: Calculate the kinetic energy of the man
The kinetic energy (KE) can be calculated using the formula:

\[ KE = \frac{1}{2} m v^2 \]

Where:
- \( m = 50 \, \text{kg} \) (mass of the man)
- \( v = 5 \, \text{m/s} \) (speed of the man)

Substituting the values:

\[ KE = \frac{1}{2} \times 50 \, \text{kg} \times (5 \, \text{m/s})^2 \]
\[ KE = \frac{1}{2} \times 50 \times 25 = 625 \, \text{J} \]

### Step 3: Determine the heat required to raise the temperature of water
We need to calculate the heat (Q) required to raise the temperature of water from \( 20^\circ C \) to \( 30^\circ C \). The formula for heat is:

\[ Q = m \cdot s \cdot \Delta T \]

Where:
- \( m \) = mass of water (in kg)
- \( s = 4200 \, \text{J/kg} \cdot \text{K} \) (specific heat capacity of water)
- \( \Delta T = 30^\circ C - 20^\circ C = 10 \, \text{K} \)

Thus, we can express Q as:

\[ Q = m \cdot 4200 \cdot 10 = 42000m \]

### Step 4: Set the kinetic energy equal to the heat required
According to the problem, the kinetic energy of the man is equal to the heat required to raise the temperature of the water:

\[ 625 \, \text{J} = 42000m \]

### Step 5: Solve for the mass of water (m)
Now, we can solve for \( m \):

\[ m = \frac{625}{42000} \]
\[ m \approx 0.01488 \, \text{kg} \]

To convert this to grams:

\[ m \approx 14.88 \, \text{g} \approx 15 \, \text{g} \]

### Final Answer
The amount of water that can be heated with the kinetic energy of the man is approximately **15 grams**.

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