A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 10⁵ 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 cm³, the change in internal energy of the sample is:
Options

Question

A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 10⁵ 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 cm³, the change in internal energy of the sample is:

Options

G1 Physics · Accepted Answer

Step 1: Use the first law of thermodynamics with work done BY the gas positive: Q = ΔU + W. Step 2: Heat absorbed to vaporise water: Q = 54 × 4.18 = 225.72 J. Step 3: The initial liquid volume is (0.1 g)/(1 g cm⁻³) = 0.1 cm³. The steam volume is 167.1 cm³. Step 4: Increase in volume: ΔV = 167.1 − 0.1 = 167.0 cm³ = 167.0 × 10⁻⁶ m³. Step 5: Work done against atmospheric pressure: W = PΔV = (1.013 × 10⁵ Pa)(167.0 × 10⁻⁶ m³) = 16.917 J. Step 6: ΔU = Q − W = 225.72 − 16.917 = 208.803 J ≈ 208.8 J. Fi…