A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 10⁵ Nm⁻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,

Question

Accepted Answer

104.3 J. The heat energy supplied (Q) is 54 cal = 54 × 4.186 J = 226.04 J. The work done (W) is PΔV = (1.013 × 10⁵) × (167.1 × 10⁻⁶) = 16.93 J. Using the first law of thermodynamics, ΔU = Q - W = 226.04 - 16.93 = 209.11 J. The closest option is 208.7 J.

A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 10⁵ Nm⁻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,

Solution

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