An experiment takes `10` minutes to raise the temperature of water in a container from `0^(@)C "to" 100^(@)C` and another `55` minutes to convert it totally into steam by a heater supplying heat at a uniform rate. Neglecting the specific heat of the container and taking specific heat of water to be `1 "cal//g ^(@)C`, the heat of vapourization according to this experiment will come out to be:-

Options

A. `540"cal"//g`
B. `530"cal"//g`
C. `560"cal"//g`
D. `550"cal"//g`

Explanation

To solve the problem, we need to calculate the heat of vaporization of water based on the given information about the heating and vaporization processes. Let's break it down step by step.

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

The formula to calculate the heat (Q) required to change the temperature of a substance is given by:
\[ Q = m \cdot s \cdot \Delta T \]
where:
- \( m \) = mass of the water (in grams)
- \( s \) = specific heat capacity of water = 1 cal/g°C
- \( \Delta T \) = change in temperature = \( 100°C - 0°C = 100°C \)

So, the heat required to raise the temperature is:
\[ Q_1 = m \cdot 1 \cdot 100 = 100m \text{ cal} \]

### Step 2: Relate the heat to the power supplied by the heater.

The power supplied by the heater can be expressed as:
\[ P = \frac{Q_1}{t_1} \]
where \( t_1 \) is the time taken to raise the temperature, which is 10 minutes (or 600 seconds).

Thus, we can write:
\[ Q_1 = P \cdot 10 \text{ minutes} = P \cdot 600 \text{ seconds} \]

### Step 3: Set up the equation for the first process.

From Step 1 and Step 2, we can equate:
\[ 100m = P \cdot 600 \]
This is our **Equation 1**.

### Step 4: Calculate the heat required to convert water at 100°C to steam at 100°C.

For the phase change from liquid to gas, we use the latent heat (L):
\[ Q_2 = m \cdot L \]
where \( L \) is the latent heat of vaporization.

### Step 5: Relate the heat for vaporization to the power supplied by the heater.

The power supplied by the heater for the vaporization process can be expressed as:
\[ P = \frac{Q_2}{t_2} \]
where \( t_2 \) is the time taken to convert water to steam, which is 55 minutes (or 3300 seconds).

Thus, we can write:
\[ Q_2 = P \cdot 55 \text{ minutes} = P \cdot 3300 \text{ seconds} \]

### Step 6: Set up the equation for the second process.

From Step 4 and Step 5, we can equate:
\[ m \cdot L = P \cdot 3300 \]
This is our **Equation 2**.

### Step 7: Divide Equation 1 by Equation 2.

Now we can divide Equation 1 by Equation 2:
\[ \frac{100m}{mL} = \frac{P \cdot 600}{P \cdot 3300} \]

Cancelling \( m \) and \( P \) from both sides gives:
\[ \frac{100}{L} = \frac{600}{3300} \]

### Step 8: Solve for the latent heat of vaporization \( L \).

Cross-multiplying gives:
\[ 100 \cdot 3300 = 600 \cdot L \]
\[ L = \frac{100 \cdot 3300}{600} \]
\[ L = \frac{330000}{600} = 550 \text{ cal/g} \]

### Conclusion

The heat of vaporization of water according to this experiment is:
\[ \boxed{550 \text{ cal/g}} \]