An electric immersion heater of `1.08 kW` is immersed in water . After it has reaches a temperature of `100^(@)C` , how much time will be required to produce `100 g` of steam?

Options

A. 50 s
B. 420 s
C. 105 s
D. 105 s

Answer

To solve the problem of how much time is required to produce 100 g of steam using an electric immersion heater of 1.08 kW, we will follow these steps:

### Step 1: Determine the heat required to produce steam
The latent heat of vaporization (L) is the amount of heat required to convert 1 gram of water at 100°C to steam at 100°C. The latent heat of vaporization for water is approximately 540 calories per gram.

For 100 grams of steam:
\[ \text{Heat required (Q)} = 100 \, \text{g} \times 540 \, \text{cal/g} = 54000 \, \text{cal} \]

### Step 2: Convert calories to joules
Since we need the heat in joules, we convert calories to joules using the conversion factor \(1 \, \text{cal} = 4.2 \, \text{J}\):
\[ Q = 54000 \, \text{cal} \times 4.2 \, \text{J/cal} = 226800 \, \text{J} \]

### Step 3: Calculate the power of the immersion heater
The power (P) of the immersion heater is given as 1.08 kW. We convert this to watts:
\[ P = 1.08 \, \text{kW} = 1.08 \times 1000 \, \text{W} = 1080 \, \text{W} \]

### Step 4: Use the formula to find time
The relationship between heat, power, and time is given by the formula:
\[ Q = P \times t \]
Rearranging the formula to find time (t):
\[ t = \frac{Q}{P} \]

### Step 5: Substitute the values
Substituting the values we have:
\[ t = \frac{226800 \, \text{J}}{1080 \, \text{W}} = 210 \, \text{s} \]

### Conclusion
The time required to produce 100 g of steam is **210 seconds**.

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