An electric room heater has a resistance of `25 Omega` and operates at `220 V` for `12` minutes. Calculate heat energy dissipated by it in kilojoules.

Explanation

To calculate the heat energy dissipated by the electric room heater, we can follow these steps:

### Step 1: Identify the given values
- Resistance (R) = 25 Ω
- Voltage (V) = 220 V
- Time (t) = 12 minutes

### Step 2: Convert time from minutes to seconds
Since time should be in seconds for our calculations:
\[ t = 12 \text{ minutes} \times 60 \text{ seconds/minute} = 720 \text{ seconds} \]

### Step 3: Calculate the power using the formula
The power (P) can be calculated using the formula:
\[ P = \frac{V^2}{R} \]
Substituting the known values:
\[ P = \frac{220^2}{25} \]

### Step 4: Calculate \(220^2\)
\[ 220^2 = 48400 \]
Now substituting this back into the power formula:
\[ P = \frac{48400}{25} = 1936 \text{ Watts} \]

### Step 5: Calculate the heat energy (Q) using the formula
The heat energy (Q) dissipated can be calculated using the formula:
\[ Q = P \times t \]
Substituting the values we have:
\[ Q = 1936 \text{ Watts} \times 720 \text{ seconds} \]

### Step 6: Calculate the total heat energy
\[ Q = 1393920 \text{ Joules} \]

### Step 7: Convert Joules to kilojoules
To convert Joules to kilojoules, we divide by 1000:
\[ Q = \frac{1393920}{1000} = 1393.92 \text{ kJ} \]

### Final Answer
The heat energy dissipated by the electric room heater is approximately **1393.92 kJ**.

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