STANDARD Physics First Law of Thermodynamics MCQ Question
1 kg of water is heated from 40°C to 70°C. If its volume remains constant, then the change in internal energy is (specific heat of water = 4148 J kg⁻¹ K⁻¹)
2.44 × 10⁵ J
1.62 × 10⁵ J
1.24 × 10⁵ J
2.62 × 10⁵ J
Correct Answer
Detailed Explanation
Since volume of water remains constant, then work done ΔW = PdV = 0. According to first law of thermodynamics dQ = dU + dW, dU = dQ = msΔT = 1 × 4148 × (70 − 40) = 4148 × 30 = 124440 J = 1.244 × 10⁵ J.
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