In a home experiment , Ram brings a new electric kettle with unknown power rating . He puts `1` litre water in the kettle and switches on . But to his dismay , the temperature becomes constants at `60^(@)C` after some time. The room temperature is `20^(@)C`. Ram gets bored and switches off the kittle . He sees that during first `20`s water cools down by `2^(@)C`.
What is the wattage of the kettle

Options

A. `840W`
B. `1W`
C. `100W`
D. `420W`

Explanation

To find the wattage of the kettle, we can follow these steps:

### Step 1: Identify the given data
- Volume of water (V) = 1 liter = 1 kg (since the density of water is approximately 1 kg/L)
- Initial temperature (T_initial) = 60°C
- Room temperature (T_room) = 20°C
- Temperature drop (ΔT) = 2°C
- Time (t) = 20 seconds

### Step 2: Calculate the heat lost by the water
The heat lost by the water can be calculated using the formula:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \( Q \) = heat lost (in joules)
- \( m \) = mass of water (in kg)
- \( c \) = specific heat capacity of water = 4200 J/(kg·°C)
- \( \Delta T \) = change in temperature (in °C)

Substituting the values:
\[ Q = 1 \, \text{kg} \cdot 4200 \, \text{J/(kg·°C)} \cdot 2 \, \text{°C} \]
\[ Q = 8400 \, \text{J} \]

### Step 3: Calculate the power of the kettle
Power (P) can be calculated using the formula:
\[ P = \frac{Q}{t} \]
where:
- \( P \) = power (in watts)
- \( Q \) = heat lost (in joules)
- \( t \) = time (in seconds)

Substituting the values:
\[ P = \frac{8400 \, \text{J}}{20 \, \text{s}} \]
\[ P = 420 \, \text{W} \]

### Final Answer
The wattage of the kettle is **420 watts**.

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