Water at `0^@C` was heated until it started to boil and then until it all changed to steam. The time required to heat water from `0^@C` to `100^@C` is 5 min and the time to change boiling water to steam is 28 minutes. If the flame supplied heat at a constant rate, the specific latent heat of vaporization of water (neglecting heat losses, container etc. is (in `J//g`). (specific heat of water s `=4.2 J//g` )

Options

A. 540
B. 2268
C. 2352
D. 2356

Explanation

To find the specific latent heat of vaporization of water, we can follow these steps:

### Step 1: Calculate the total heat supplied to raise the temperature of water from 0°C to 100°C.

The formula for heat supplied (Q) is given by:
\[ Q = m \cdot s \cdot \Delta T \]
Where:
- \( m \) = mass of water (in grams)
- \( s \) = specific heat of water = 4.2 J/g°C
- \( \Delta T \) = change in temperature = \( 100°C - 0°C = 100°C \)

So, the heat supplied to raise the temperature from 0°C to 100°C is:
\[ Q_1 = m \cdot 4.2 \cdot 100 \]
\[ Q_1 = 420m \, \text{J} \]

### Step 2: Calculate the total heat supplied to convert boiling water to steam.

The heat required to convert boiling water to steam is given by:
\[ Q = m \cdot L \]
Where:
- \( L \) = specific latent heat of vaporization (in J/g)

The time taken to convert boiling water to steam is 28 minutes, which is:
\[ t_2 = 28 \times 60 = 1680 \, \text{s} \]

The heat supplied during this time is:
\[ Q_2 = P \cdot t_2 \]
Where \( P \) is the power (rate of heat supply).

### Step 3: Relate the power to the heat supplied.

The power can also be expressed in terms of the heat supplied to raise the temperature:
\[ P = \frac{Q_1}{t_1} \]
Where \( t_1 = 5 \times 60 = 300 \, \text{s} \) (time to heat water from 0°C to 100°C).

Thus:
\[ P = \frac{420m}{300} \]
\[ P = 1.4m \, \text{W} \]

### Step 4: Substitute power into the heat equation for boiling water to steam.

Using the expression for power in the heat equation for boiling water to steam:
\[ Q_2 = P \cdot t_2 \]
\[ Q_2 = (1.4m) \cdot (1680) \]
\[ Q_2 = 2352m \, \text{J} \]

### Step 5: Set the two expressions for heat equal to each other.

From the heat required to convert boiling water to steam:
\[ Q_2 = m \cdot L \]
So we have:
\[ 2352m = m \cdot L \]

### Step 6: Solve for the specific latent heat of vaporization (L).

Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ L = 2352 \, \text{J/g} \]

### Final Answer:
The specific latent heat of vaporization of water is \( 2352 \, \text{J/g} \).

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