STANDARD Physics First Law of Thermodynamics MCQ 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 into steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is
104.3 J
208.7 J
42.2 J
84.5 J
Correct Answer
Detailed Explanation
Q = 54 cal = 54 × 4.18 joule = 225.72 joule. W = P(Vₛₜₑₐₘ − Vₜₐₜₑᵣ). For water 0.1 g = 0.1 cc. W = 1.013 × 10⁵ [167 × 10⁻⁶ − 0.1 × 10⁻⁶] joule = 16.917 joule. By first law of thermodynamics, U = ΔQ − ΔW = 225.72 − 16.97 = 208.8 Joule.
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