If a given volume of water in a `220 V` heater is boiled in `5 min`, then how much time will it take for the same volume of water in a `110 V` heater to be boiled?

Options

A. `20 min`
B. `30 min `
C. `25 min`
D. `40 min`

Explanation

To solve the problem of how much time it will take to boil the same volume of water in a 110 V heater, given that it takes 5 minutes in a 220 V heater, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Heat Required for Boiling**:
The amount of heat required to boil a given mass of water is constant and can be expressed as:
\[ Q = m \cdot L \]
where \( Q \) is the heat required, \( m \) is the mass of the water, and \( L \) is the latent heat of vaporization.

2. **Heat Generated by the Heaters**:
The heat generated by an electric heater can be expressed using the formula:
\[ H = \frac{V^2}{R} \cdot t \]
where \( H \) is the heat generated, \( V \) is the voltage, \( R \) is the resistance of the heater, and \( t \) is the time.

3. **Set Up the Equations for Both Heaters**:
For the 220 V heater, the heat generated in 5 minutes (which is 300 seconds) can be expressed as:
\[ H_1 = \frac{(220)^2}{R} \cdot 300 \]
For the 110 V heater, the heat generated in time \( t_2 \) can be expressed as:
\[ H_2 = \frac{(110)^2}{R} \cdot t_2 \]

4. **Equate the Heat Generated**:
Since the heat required to boil the water is the same for both heaters, we can set \( H_1 = H_2 \):
\[ \frac{(220)^2}{R} \cdot 300 = \frac{(110)^2}{R} \cdot t_2 \]

5. **Simplify the Equation**:
The resistance \( R \) cancels out from both sides:
\[ (220)^2 \cdot 300 = (110)^2 \cdot t_2 \]
Now, substituting the values:
\[ 48400 \cdot 300 = 12100 \cdot t_2 \]

6. **Solve for \( t_2 \)**:
Rearranging the equation gives:
\[ t_2 = \frac{48400 \cdot 300}{12100} \]
Simplifying the right side:
\[ t_2 = \frac{48400}{12100} \cdot 300 = 4 \cdot 300 = 1200 \text{ seconds} \]
Converting seconds to minutes:
\[ t_2 = \frac{1200}{60} = 20 \text{ minutes} \]

### Final Answer:
It will take **20 minutes** to boil the same volume of water in a 110 V heater.