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QPages 62-64: Q19: HC Verma: Concepts of Physics - Volume 2 Laws of Thermodynamics Pages 62-64: Q19 Solution: Find the change in the internal energy of 2 kg of water as it…

Find the change in the internal energy of 2 kg of water as it is heated from 0�C to 4�C. The specific heat capacity of water is 4200 J kg−1 K−14200 \text{ J kg}^{-1} \text{ K}^{-1} and its densities at 0�C and 4�C are 999.9 kg m−3999.9 \text{ kg m}^{-3} and 1000 kg m−31000 \text{ kg m}^{-3} respectively. Atmospheric pressure = 105 Pa10^5 \text{ Pa}.

Step-by-Step Verified Solution

**Answer:** ΔU=33600\Delta U = 33600 J. **Step-by-Step Explanation:** Given: m=2m = 2 kg, ΔT=4\Delta T = 4 K, c=4200c = 4200 J/kg-K. ΔU=mcΔT=2×4200×4=33600\Delta U = mc\Delta T = 2 \times 4200 \times 4 = 33600 J.

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