A sample of `0.1 g` of water of `100^(@)C` and normal pressure `(1.013 xx 10^(5) N m^(-2))` requires 54 cal of heat energy to convert to steam at `100^(@)C`. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample is

Options

A. 42.2J
B. 208.7J
C. 104.3J
D. 84.5J

Answer

To find the change in internal energy of the water sample as it converts to steam, we can follow these steps:

### Step 1: Understand the relationship between heat, work done, and change in internal energy.
The change in internal energy (ΔU) can be expressed as:
\[ \Delta U = Q - W \]
where \(Q\) is the heat added to the system and \(W\) is the work done by the system.

### Step 2: Identify the heat added (Q).
From the problem, we know that the heat required to convert the water to steam is \(54 \, \text{cal}\). We need to convert this to joules since the work done will be calculated in joules. The conversion factor is:
\[ 1 \, \text{cal} = 4.18 \, \text{J} \]
Thus,
\[ Q = 54 \, \text{cal} \times 4.18 \, \text{J/cal} = 226.92 \, \text{J} \]

### Step 3: Calculate the work done (W).
The work done during the phase change can be calculated using the formula:
\[ W = P \Delta V \]
where \(P\) is the pressure and \(\Delta V\) is the change in volume.

Given:
- The pressure \(P = 1.013 \times 10^5 \, \text{N/m}^2\)
- The final volume of steam produced is \(167.1 \, \text{cc} = 167.1 \times 10^{-6} \, \text{m}^3\)
- The initial volume of water (0.1 g) can be calculated using the density of water (1 g/cc):
\[ \text{Initial volume} = \frac{0.1 \, \text{g}}{1 \, \text{g/cc}} = 0.1 \, \text{cc} = 0.1 \times 10^{-6} \, \text{m}^3 \]

Now, calculate \(\Delta V\):
\[ \Delta V = V_{\text{final}} - V_{\text{initial}} = (167.1 \times 10^{-6} \, \text{m}^3) - (0.1 \times 10^{-6} \, \text{m}^3) = 167.0 \times 10^{-6} \, \text{m}^3 \]

Now, calculate the work done:
\[ W = P \Delta V = (1.013 \times 10^5 \, \text{N/m}^2)(167.0 \times 10^{-6} \, \text{m}^3) = 16.917 \, \text{J} \]

### Step 4: Calculate the change in internal energy (ΔU).
Now we can substitute the values of \(Q\) and \(W\) into the equation for ΔU:
\[ \Delta U = Q - W = 226.92 \, \text{J} - 16.917 \, \text{J} = 209.003 \, \text{J} \]

### Final Answer:
The change in internal energy of the sample is approximately:
\[ \Delta U \approx 209.003 \, \text{J} \]