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QPages 12-14: Q17: HC Verma: Concepts of Physics - Volume 2 Heat and Temperature Pages 12-14: Q17 Solution: The density of water at 0�C is and at 4�C is . Calculate the…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the direct linear expansion formula with density, failing to recognize that density is inversely proportional to volume ($\rho \propto 1/V$). To avoid this, use the specific density relation $\gamma = \frac{\rho_2 - \rho_1}{\rho_1 \Delta T}$ directly, rather than attempting to apply the standard $1 + \gamma \Delta T$ volume expansion factor to density values.

The density of water at 0�C is 0.998 g cm−30.998 \text{ g cm}^{-3} and at 4�C is 1.000 g cm−31.000 \text{ g cm}^{-3}. Calculate the average coefficient of volume expansion of water in the temperature range of 0 to 4�C.

Step-by-Step Verified Solution

**Answer:** γavg=1.25×10−4\gamma_{avg} = 1.25 \times 10^{-4} /�C **Step-by-Step Explanation:** Density at 0�C: ρ0=0.998\rho_0 = 0.998 g/cm_. At 4�C: ρ4=1.000\rho_4 = 1.000 g/cm_. Using ρ4=ρ0(1−γΔT)\rho_4 = \rho_0(1 - \gamma \Delta T) with ΔT=4\Delta T = 4:
1.000=0.998(1−4γ)⇒1−4γ=1.00200⇒4γ=−0.00200⇒γ=−5×10−41.000 = 0.998(1 - 4\gamma) \Rightarrow 1 - 4\gamma = 1.00200 \Rightarrow 4\gamma = -0.00200 \Rightarrow \gamma = -5 \times 10^{-4}
Wait � the density increases from 0 to 4�C (anomalous expansion). Average coefficient:
γ=ρ4−ρ0ρ0ΔT=1.000−0.9980.998×4=5.01×10−4\gamma = \frac{\rho_4 - \rho_0}{\rho_0 \Delta T} = \frac{1.000 - 0.998}{0.998 \times 4} = 5.01 \times 10^{-4}

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