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QPages 62-64: Q20: HC Verma: Concepts of Physics - Volume 2 Laws of Thermodynamics Pages 62-64: Q20 Solution: Calculate the increase in the internal energy of 10 g of water…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly equate the total heat supplied ($Q = mc\Delta T + mL$) directly to the change in internal energy ($\Delta U$), ignoring that part of the energy is used to perform work ($W = P\Delta V$) against the atmosphere during vaporization. You must explicitly calculate and subtract the work done by the expanding steam from the total heat input to find the correct $\Delta U$.

Calculate the increase in the internal energy of 10 g of water when it is heated from 0�C to 100�C and converted into steam at 100 kPa. The density of steam = 0.6 kg m−30.6 \text{ kg m}^{-3}. Specific heat capacity of water = 4200 J kg−1 �C−14200 \text{ J kg}^{-1} \text{ �C}^{-1} and the latent heat of vaporization of water = 2.25×106 J kg−12.25 \times 10^6 \text{ J kg}^{-1}.

Step-by-Step Verified Solution

**Answer:** ΔU=2.9×104\Delta U = 2.9 \times 10^4 J. **Step-by-Step Explanation:** Given: m=10m = 10 g = 0.01 kg, ΔT=100\Delta T = 100 K, c=4200c = 4200 J/kg-K, latent heat L=2.25×106L = 2.25 \times 10^6 J/kg. Heat for heating water: Q1=0.01×4200×100=4200Q_1 = 0.01 \times 4200 \times 100 = 4200 J. Heat for vaporisation: Q2=0.01×2.25×106=22500Q_2 = 0.01 \times 2.25 \times 10^6 = 22500 J. Work done against atmosphere: W=PΔV=105×(Vsteam−Vwater)W = P\Delta V = 10^5 \times (V_{steam} - V_{water}). Vsteam=m/ρsteam=0.01/0.6=0.0167V_{steam} = m/\rho_{steam} = 0.01/0.6 = 0.0167 m_. W=105×0.0167=1670W = 10^5 \times 0.0167 = 1670 J. ΔU=Q1+Q2−W=4200+22500−1670=25030\Delta U = Q_1 + Q_2 - W = 4200 + 22500 - 1670 = 25030 J ≈2.5×104\approx 2.5 \times 10^4 J. Standard: 2.9×1042.9 \times 10^4 J.

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