An immersion heater, in an insulated vessel of negligible heat capacity brings 100 g of water to the boiling point from `16^@C` in 7 min. Then
Q. Power of heat is nearly

Options

A. `8.4xx10^3`
B. 84 W
C. `8.4xx10^3(cal)/(s)`
D. 20 W

Explanation

To find the power of the immersion heater that brings 100 g of water to the boiling point from 16°C in 7 minutes, we can follow these steps:

### Step 1: Identify the given values
- Mass of water (m) = 100 g = 100 × 10^(-3) kg = 0.1 kg
- Initial temperature (T1) = 16°C
- Final temperature (T2) = 100°C
- Time (t) = 7 minutes = 7 × 60 seconds = 420 seconds
- Specific heat capacity of water (c) = 4184 J/(kg·°C)

### Step 2: Calculate the change in temperature (ΔT)
\[ \Delta T = T2 - T1 = 100°C - 16°C = 84°C \]

### Step 3: Calculate the heat (Q) required to raise the temperature of the water
Using the formula:
\[ Q = mc\Delta T \]
Substituting the known values:
\[ Q = (0.1 \, \text{kg}) \times (4184 \, \text{J/(kg·°C)}) \times (84 \, \text{°C}) \]
\[ Q = 0.1 \times 4184 \times 84 \]
\[ Q = 0.1 \times 351456 = 35145.6 \, \text{J} \]

### Step 4: Calculate the power (P) of the heater
Power is defined as the rate of energy transfer, given by:
\[ P = \frac{Q}{t} \]
Substituting the values we have:
\[ P = \frac{35145.6 \, \text{J}}{420 \, \text{s}} \]
\[ P \approx 83.6 \, \text{W} \]

### Step 5: Round the answer
Since we are asked for the power nearly, we can round it to:
\[ P \approx 84 \, \text{W} \]

### Final Answer
The power of the immersion heater is approximately **84 Watts**.

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