How much heat energy is gained when 5 kg of water at `20^(@)C` is brought to its boiling point (Specific heat of water `= 4.2` kj kg c)

Options

A. 1680 kJ
B. 1700 kJ
C. 1720 kJ
D. 1740 kJ

Explanation

To find the heat energy gained when 5 kg of water at 20°C is brought to its boiling point, we can use the formula for heat energy:

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

Where:
- \( Q \) = heat energy (in kilojoules)
- \( m \) = mass of the water (in kg)
- \( s \) = specific heat capacity of water (in kJ/kg°C)
- \( \Delta T \) = change in temperature (in °C)

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of water, \( m = 5 \, \text{kg} \)
- Initial temperature, \( T_i = 20 \, \text{°C} \)
- Boiling point of water, \( T_f = 100 \, \text{°C} \)
- Specific heat of water, \( s = 4.2 \, \text{kJ/kg°C} \)

2. **Calculate the change in temperature (\( \Delta T \)):**
\[ \Delta T = T_f - T_i = 100 \, \text{°C} - 20 \, \text{°C} = 80 \, \text{°C} \]

3. **Substitute the values into the heat energy formula:**
\[ Q = m \cdot s \cdot \Delta T \]
\[ Q = 5 \, \text{kg} \cdot 4.2 \, \text{kJ/kg°C} \cdot 80 \, \text{°C} \]

4. **Perform the multiplication:**
- First, calculate \( 5 \cdot 80 = 400 \)
- Then, calculate \( 400 \cdot 4.2 = 1680 \)

5. **Final result:**
\[ Q = 1680 \, \text{kJ} \]

Thus, the heat energy gained when 5 kg of water at 20°C is brought to its boiling point is **1680 kJ**.