An electric kettle takes `4 A` current at `220 V`. How much time will it take to boil `1 kg` of wate from temperature `20^(@)C`? The temperature of boiling water is `100^(@)C`

Options

A. `6.3 min`
B. `8.4 min`
C. `12.6 min`
D. `4.2 min`

Explanation

To solve the problem of how long it will take to boil 1 kg of water from 20°C to 100°C using an electric kettle that takes 4 A current at 220 V, we can follow these steps:

### Step 1: Calculate the power of the electric kettle
The power (P) generated by the kettle can be calculated using the formula:
\[ P = V \times I \]
Where:
- \( V = 220 \, V \) (voltage)
- \( I = 4 \, A \) (current)

Substituting the values:
\[ P = 220 \, V \times 4 \, A = 880 \, W \]
This means the kettle generates 880 joules of heat per second.

### Step 2: Calculate the heat required to boil the water
The heat (Q) required to raise the temperature of the water can be calculated using the formula:
\[ Q = m \times C \times \Delta T \]
Where:
- \( m = 1 \, kg \) (mass of water)
- \( C = 4200 \, \text{J/kg°C} \) (specific heat capacity of water)
- \( \Delta T = 100°C - 20°C = 80°C \) (temperature change)

Substituting the values:
\[ Q = 1 \, kg \times 4200 \, \text{J/kg°C} \times 80°C \]
\[ Q = 1 \times 4200 \times 80 = 336000 \, J \]

### Step 3: Calculate the time required to boil the water
The time (t) required to heat the water can be calculated using the formula:
\[ t = \frac{Q}{P} \]
Substituting the values:
\[ t = \frac{336000 \, J}{880 \, W} \]
\[ t = \frac{336000}{880} \approx 381.82 \, seconds \]

### Step 4: Convert time from seconds to minutes
To convert seconds to minutes, divide by 60:
\[ t \approx \frac{381.82}{60} \approx 6.36 \, minutes \]

### Final Answer
It will take approximately **6.36 minutes** to boil 1 kg of water from 20°C to 100°C.

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