A `50 kg` man is running at a speed of `18 kmh^(-1)` If all the kinetic energy of the man can be used to increase the temperature of water from `20^(@)C` to `30^(@)C`.How much water can be heated with this energy?

Explanation

To solve the problem, we will follow these steps:

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

### Step 2: Calculate the kinetic energy of the man
The formula for kinetic energy (KE) is:
\[ KE = \frac{1}{2} mv^2 \]
Where:
- \( m = 50 \text{ kg} \) (mass of the man)
- \( v = 5 \text{ m/s} \) (speed of the man)

Substituting the values:
\[ KE = \frac{1}{2} \times 50 \times (5)^2 = \frac{1}{2} \times 50 \times 25 = 625 \text{ joules} \]

### Step 3: Determine the heat required to raise the temperature of water
The heat added to the water can be calculated using the formula:
\[ q = mc\Delta T \]
Where:
- \( m \) = mass of water (in kg)
- \( c \) = specific heat capacity of water = \( 4200 \text{ J/(kg°C)} \)
- \( \Delta T = T_f - T_i = 30°C - 20°C = 10°C \)

We can set the heat added equal to the kinetic energy:
\[ 625 = m \cdot 4200 \cdot 10 \]

### Step 4: Solve for the mass of water
Rearranging the equation to find \( m \):
\[ m = \frac{625}{4200 \cdot 10} \]
Calculating this gives:
\[ m = \frac{625}{42000} \approx 0.01488 \text{ kg} \]

### Step 5: Convert the mass of water to grams (if necessary)
Since \( 1 \text{ kg} = 1000 \text{ grams} \):
\[ m \approx 0.01488 \times 1000 = 14.88 \text{ grams} \]

### Final Answer
The amount of water that can be heated from \( 20°C \) to \( 30°C \) using the kinetic energy of the man is approximately \( 0.01488 \text{ kg} \) or \( 14.88 \text{ grams} \).

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