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. Let's break it down step by step.

### Step 1: Understand the Problem
We need to find the heat of vaporization of water using the time taken to heat water from 0°C to 100°C and then convert it to steam.

### Step 2: Calculate the Heat Required to Raise the Temperature
The formula for heat (Q) required to raise the temperature of a substance is given by:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \( m \) = mass of the water (in grams)
- \( c \) = specific heat of water (1 cal/g°C)
- \( \Delta T \) = change in temperature (final temperature - initial temperature)

In this case:
- \( \Delta T = 100°C - 0°C = 100°C \)

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

### Step 3: Calculate the Heat Required for Vaporization
The heat required to convert water at 100°C to steam is given by:
\[ Q = m \cdot L_v \]
where:
- \( L_v \) = latent heat of vaporization (cal/g)

From the problem, we know that this process takes 55 minutes. The power supplied by the heater is constant, so we can express the heat supplied as:
\[ Q_2 = P \cdot t \]
where \( P \) is the power and \( t \) is the time in minutes.

### Step 4: Relate the Two Heat Equations
Since the power is constant, we can set up the following relationship:
1. Heat to raise the temperature:
\[ Q_1 = P \cdot 10 \text{ minutes} \]
2. Heat to vaporize the water:
\[ Q_2 = P \cdot 55 \text{ minutes} \]

### Step 5: Set Up the Equation
From the above, we can write:
\[ 100m = P \cdot 10 \]
\[ m \cdot L_v = P \cdot 55 \]

### Step 6: Divide the Two Equations
Now, we can divide the two equations to eliminate \( P \):
\[ \frac{100m}{m \cdot L_v} = \frac{P \cdot 10}{P \cdot 55} \]

This simplifies to:
\[ \frac{100}{L_v} = \frac{10}{55} \]

### Step 7: Solve for \( L_v \)
Cross-multiplying gives:
\[ 100 \cdot 55 = 10 \cdot L_v \]
\[ 5500 = 10 \cdot L_v \]
\[ L_v = \frac{5500}{10} = 550 \text{ cal/g} \]

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