At `0^(@)C` and normal atmospheric pressure, the volume of 1 gram of water increases from `1 c.c "to" 1.091 c.c` on freezing. What will be the change in its internal energy? Normal atmospheric pressure is `1.013xx10^(5)N//m^(2)` and latent heat of melting of ice `= 80 cal//gram`.

Explanation

To find the change in internal energy (ΔU) when 1 gram of water freezes at 0°C, we can follow these steps:

### Step 1: Identify the given values
- Mass of water (m) = 1 gram
- Change in volume (ΔV) = 1.091 cm³ - 1 cm³ = 0.091 cm³
- Normal atmospheric pressure (P) = 1.013 x 10^5 N/m²
- Latent heat of melting of ice (L) = 80 cal/gram

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