To calculate the change in internal energy during the freezing of water, we need to consider the specific heat capacity and the change in volume.
Given: Mass of water (m) = 1 gram Initial volume (V₁) = 1 cm³ Final volume (V₂) = 1.05 cm³ Total heat of fusion (ΔH) = 80 cal/g
First, let's convert the units:
1 cal = 4.186 J (Joules) 1 atm = 101,325 Pa (Pascals) 1 Pa = 1 N/m² (Newtons per square meter)
So, 80 cal/g = 80 × 4.186 J/g ≈ 334.88 J/g
Next, we need to calculate the change in volume (ΔV):
ΔV = V₂ - V₁ = 1.05 cm³ - 1 cm³ = 0.05 cm³
Now, we can calculate the change in internal energy (ΔU):
ΔU = ΔH + PΔV
Here, P is the pressure, and at standard temperature and pressure (STP), the pressure is 1 atm, which is approximately 101,325 Pa.
Converting P to N/m²:
P = 101,325 Pa = 101,325 N/m²
ΔU = ΔH + PΔV = 334.88 J/g + (101,325 N/m²) × (0.05 cm³) × (1 × 10^(-6) m³/cm³) = 334.88 J/g + 5.06625 J/g = 339.94625 J/g
Since 1 ATP (atmosphere pressure) is equal to 1.013 × 10^5 N/m², we can convert the change in internal energy to ATP:
ΔU = 339.94625 J/g × (1 ATP / 1.013 × 10^5 J/g) ≈ 3.355 × 10^(-3) ATP
Therefore, the change in internal energy during the freezing of water is approximately 3.355 × 10^(-3) ATP.