It takes 10 min for an electric kettle to heat a certain quantity of water from `0^@C` to `100^@C`. It takes 54 min to convert this water `100^@C` into steam Then latent heat of steam is

Options

A. 80 cal/g
B. 540 cal/kg
C. 540 cal/g
D. 80 cal/kg

Explanation

To find the latent heat of steam, we can follow these steps:

### Step 1: Understand the problem
We need to calculate the latent heat of steam given that it takes 10 minutes to heat water from 0°C to 100°C and 54 minutes to convert that water at 100°C into steam.

### Step 2: Calculate the heat absorbed to heat water from 0°C to 100°C
The formula to calculate the heat absorbed (Q) when heating a substance is:
\[ Q = m \cdot S \cdot \Delta T \]
Where:
- \( m \) = mass of water (in grams)
- \( S \) = specific heat capacity of water (1 cal/g°C)
- \( \Delta T \) = change in temperature (100°C - 0°C = 100°C)

Substituting the values:
\[ Q_1 = m \cdot 1 \cdot (100 - 0) = 100m \, \text{calories} \]

### Step 3: Calculate the heat absorbed to convert water at 100°C to steam
The heat absorbed to convert water into steam is given by:
\[ Q_2 = m \cdot L \]
Where:
- \( L \) = latent heat of steam (in cal/g)

### Step 4: Relate the time taken to the heat absorbed
The time taken to heat the water from 0°C to 100°C is 10 minutes, and the time taken to convert the water to steam is 54 minutes. The heat absorbed in 54 minutes can be expressed as:
\[ Q_2 = Q_1 \cdot \frac{54 \, \text{min}}{10 \, \text{min}} = 100m \cdot \frac{54}{10} = 540m \, \text{calories} \]

### Step 5: Set the equations equal to each other
Since both expressions represent the heat absorbed during the conversion of water to steam, we can set them equal:
\[ m \cdot L = 540m \]

### Step 6: Solve for the latent heat \( L \)
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ L = 540 \, \text{calories/g} \]

### Conclusion
The latent heat of steam is \( 540 \, \text{calories/g} \).