Problem 2.149 — Liquid Surface: Temperature Dependence of $\sigma$

Problem Statement

Solve the thermodynamics problem: Surface tension typically decreases with temperature. For water: $\sigma(T) \approx \sigma_0(1-T/T_c)^\mu$ with $\mu\approx1.26$ and $T_c=647\ \text{K}$. At $T=20°\text{C}$, estimate $\sigma$. (Given $\sigma_0\approx0.235\ \text{N/m}$) All quantities, constants, and constraints stated in the problem

Given Information

  • See problem statement for all given quantities.

Physical Concepts & Formulas

Thermodynamics governs energy transformations involving heat and work. The First Law $\Delta U = Q – W$ expresses energy conservation. For an ideal gas, internal energy depends only on temperature ($U = nC_VT$), and the equation of state $PV = nRT$ links pressure, volume, and temperature.

  • $\Delta U = Q – W$ — First Law of Thermodynamics
  • $PV = nRT$ — ideal gas equation
  • $C_P – C_V = R$, $\gamma = C_P/C_V$
  • $W = \int P\,dV$ — work done by gas

Step-by-Step Solution

Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Worked Calculation

$$\eta = 1 – \frac{300}{600} = 1 – 0.5 = 0.50 = 50\%$$

$$\boxed{\eta_{\text{Carnot}} = 1 – \dfrac{T_C}{T_H}}$$

Answer

$$\boxed{\eta_{\text{Carnot}} = 1 – \dfrac{T_C}{T_H}}$$

Physical Interpretation

The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.


Comments

Leave a Reply

Your email address will not be published. Required fields are marked *