At 1 atmospheric pressure, 1000 g of water having a volume of `1000 cm^3` becomes `1.097 cm^3` of ice on freezing. The heat of fusion of water at 1 atmosphere is `80.0 cal//g`. What is the change in internal energy during the process?

Explanation

To find the change in internal energy during the process of freezing water into ice, we will follow these steps:

### Step 1: Identify the given data
- Mass of water, \( m = 1000 \, \text{g} \)
- Heat of fusion of water, \( L = 80.0 \, \text{cal/g} \)
- Initial volume of water, \( V_{\text{water}} = 1000 \, \text{cm}^3 \)
- Final volume of ice, \( V_{\text{ice}} = 1.097 \, \text{cm}^3 \)
- Pressure, \( P = 1 \, \text{atm} = 10^5 \, \text{Pa} \)

### Step 2: Calculate the heat removed during the freezing process
The heat removed during the freezing process can be calculated using the formula:
\[ \Delta Q = -mL \]
Substituting the values:
\[ \Delta Q = -1000 \, \text{g} \times 80.0 \, \text{cal/g} = -80000 \, \text{cal} \]

### Step 3: Calculate the change in volume
The change in volume, \( \Delta V \), is given by:
\[ \Delta V = V_{\text{ice}} - V_{\text{water}} = 1.097 \, \text{cm}^3 - 1000 \, \text{cm}^3 = -998.903 \, \text{cm}^3 \]

### Step 4: Calculate the work done on the system
The work done by the system can be calculated using the formula:
\[ W = P \Delta V \]
First, convert \( \Delta V \) to cubic meters:
\[ \Delta V = -998.903 \, \text{cm}^3 = -998.903 \times 10^{-6} \, \text{m}^3 \]
Now, substituting the values:
\[ W = 10^5 \, \text{Pa} \times (-998.903 \times 10^{-6} \, \text{m}^3) \approx -99.8903 \, \text{J} \]
To convert joules to calories (using \( 1 \, \text{cal} = 4.184 \, \text{J} \)):
\[ W \approx -99.8903 \, \text{J} \times \frac{1 \, \text{cal}}{4.184 \, \text{J}} \approx -23.9 \, \text{cal} \]

### Step 5: Calculate the change in internal energy
Using the first law of thermodynamics:
\[ \Delta U = \Delta Q - W \]
Substituting the values:
\[ \Delta U = -80000 \, \text{cal} - (-23.9 \, \text{cal}) = -80000 + 23.9 \approx -79976.1 \, \text{cal} \]

### Final Answer
The change in internal energy during the freezing process is approximately:
\[ \Delta U \approx -79976.1 \, \text{cal} \]