1 cc of water becomes 1681 cc of steam when boiled at a pressure of `10^5 Nm^2`. The increasing internal energy of the system is (L.T. of steam is 540 cal `g ^(-1)`, 1 calorie = 4.2J)

Options

A. 300 cal
B. 500 cal
C. 225 cal
D. 600 cal

Explanation

To solve the problem, we need to calculate the increasing internal energy of the system when 1 cc of water is converted to 1681 cc of steam at a pressure of \(10^5 \, \text{N/m}^2\). The latent heat of steam is given as \(540 \, \text{cal/g}\), and we know that \(1 \, \text{cal} = 4.2 \, \text{J}\).

### Step-by-Step Solution:

1. **Calculate the Mass of Water**:
- Given that the volume of water is \(1 \, \text{cc}\) (which is equivalent to \(1 \, \text{g}\) since the density of water is \(1 \, \text{g/cc}\)).
\[ m = 1 \, \text{g} \]

2. **Calculate the Heat Required (dq)**:
- The heat required to convert water to steam is given by the formula:
\[ dq = m \times L \]
- Where \(L\) is the latent heat of steam.
- Substituting the values:
\[ dq = 1 \, \text{g} \times 540 \, \text{cal/g} = 540 \, \text{cal} \]
- Convert calories to joules:
\[ dq = 540 \, \text{cal} \times 4.2 \, \text{J/cal} = 2268 \, \text{J} \]

3. **Calculate the Work Done (dW)**:
- The work done during the phase change can be calculated using:
\[ dW = P \times \Delta V \]
- Where \(P\) is the pressure and \(\Delta V\) is the change in volume.
- The initial volume \(V_1 = 1 \, \text{cc} = 1 \times 10^{-6} \, \text{m}^3\) and the final volume \(V_2 = 1681 \, \text{cc} = 1681 \times 10^{-6} \, \text{m}^3\).
- Therefore, the change in volume is:
\[ \Delta V = V_2 - V_1 = (1681 - 1) \times 10^{-6} \, \text{m}^3 = 1680 \times 10^{-6} \, \text{m}^3 \]
- Now substituting the values:
\[ dW = 10^5 \, \text{N/m}^2 \times 1680 \times 10^{-6} \, \text{m}^3 = 168.0 \, \text{J} \]

4. **Calculate the Change in Internal Energy (dU)**:
- According to the first law of thermodynamics:
\[ dq = dU + dW \]
- Rearranging gives us:
\[ dU = dq - dW \]
- Substituting the values we calculated:
\[ dU = 2268 \, \text{J} - 168.0 \, \text{J} = 2100 \, \text{J} \]

5. **Convert the Change in Internal Energy to Calories**:
- To convert joules to calories:
\[ dU = \frac{2100 \, \text{J}}{4.2 \, \text{J/cal}} \approx 500 \, \text{cal} \]

### Final Answer:
The increasing internal energy of the system is \(2100 \, \text{J}\) or \(500 \, \text{cal}\).