1 g of water at `100^(@)C` is completely converted into steam at `100^(@)C`. 1g of steam occupies a volume of 1650cc. (Neglect the volume of 1g of water at `100^(@)C`). At the pressure of `10^5N//m^2`, latent heat of steam is 540 cal/g (1 Calorie=4.2 joules). The increase in the internal energy in joules is

Options

A. 2310
B. 2103
C. 1650
D. 2150

Answer

To find the increase in internal energy when 1 g of water at 100°C is converted into steam at the same temperature, we can follow these steps:

### Step 1: Calculate the heat given to the system (Q)
The heat required to convert water to steam is given by the formula:
\[ Q = m \times L \]
where:
- \( m \) = mass of water (1 g)
- \( L \) = latent heat of vaporization (540 cal/g)

First, we need to convert the latent heat from calories to joules:
\[ 1 \text{ Cal} = 4.2 \text{ Joules} \]
Thus,
\[ L = 540 \text{ cal/g} \times 4.2 \text{ J/Cal} = 2268 \text{ J/g} \]

Now, substituting the values:
\[ Q = 1 \text{ g} \times 2268 \text{ J/g} = 2268 \text{ J} \]

### Step 2: Calculate the work done by the system (W)
The work done during the expansion at constant pressure can be calculated using:
\[ W = P \Delta V \]
where:
- \( P = 10^5 \text{ N/m}^2 \) (pressure)
- \( \Delta V = V_{final} - V_{initial} \)

Given that the volume of 1 g of steam is 1650 cc, we convert this to cubic meters:
\[ V_{final} = 1650 \text{ cc} = 1650 \times 10^{-6} \text{ m}^3 = 1.65 \times 10^{-3} \text{ m}^3 \]
Since we neglect the volume of water, we can assume:
\[ V_{initial} = 0 \text{ m}^3 \]

Thus,
\[ \Delta V = 1.65 \times 10^{-3} \text{ m}^3 - 0 = 1.65 \times 10^{-3} \text{ m}^3 \]

Now, substituting the values into the work formula:
\[ W = 10^5 \text{ N/m}^2 \times 1.65 \times 10^{-3} \text{ m}^3 = 165 \text{ J} \]

### Step 3: Calculate the increase in internal energy (ΔU)
Using the first law of thermodynamics:
\[ \Delta U = Q - W \]
Substituting the values we calculated:
\[ \Delta U = 2268 \text{ J} - 165 \text{ J} = 2103 \text{ J} \]

### Final Answer
The increase in internal energy is:
\[ \Delta U = 2103 \text{ J} \]

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