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`

Answer

To solve the problem of how much heat energy is gained when 5 kg of water at 20°C is brought to its boiling point (100°C), we can use the formula for heat energy gained, which is:

\[ Q = M \times S \times \Delta T \]

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

### Step 1: Identify the values
- Mass of water, \( M = 5 \) kg
- Specific heat of water, \( S = 4.2 \) kJ/kg°C
- Initial temperature, \( T_i = 20 \)°C
- Final temperature, \( T_f = 100 \)°C

### Step 2: Calculate the change in temperature (\( \Delta T \))
\[ \Delta T = T_f - T_i = 100°C - 20°C = 80°C \]

### Step 3: Substitute the values into the formula
Now we can substitute the values into the heat energy formula:
\[ Q = M \times S \times \Delta T \]
\[ Q = 5 \, \text{kg} \times 4.2 \, \text{kJ/kg°C} \times 80 \, \text{°C} \]

### Step 4: Perform the multiplication
\[ Q = 5 \times 4.2 \times 80 \]
Calculating step-by-step:
1. First, calculate \( 5 \times 80 = 400 \)
2. Then, calculate \( 400 \times 4.2 = 1680 \)

### Step 5: Conclusion
Thus, the total heat energy gained when 5 kg of water is heated from 20°C to 100°C is:
\[ Q = 1680 \, \text{kJ} \]

### Final Answer
The heat energy gained is **1680 kJ**.

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