Problem Statement
The density of mercury at 0°C is $13600$ kg/m³. Find its density at 100°C. ($\gamma_{Hg} = 1.8 \times 10^{-4}$ °C$^{-1}$)
Given Information
- All quantities, constants, and constraints stated in the problem above
- Physical constants used as needed (see Concepts section)
Physical Concepts & Formulas
This problem draws on fundamental physical principles. The key is to identify which conservation law or field equation governs the system, then apply it systematically. Dimensional analysis can always be used to verify that the final answer has the correct units. Working from first principles — rather than memorising formulas — builds deeper understanding and allows tackling novel problems.
- Identify the relevant physical law (Newton’s laws, conservation of energy/momentum, Maxwell’s equations, etc.)
- State the mathematical form of that law as it applies here
- Check dimensions at every step: both sides of an equation must have the same units
Step-by-Step Solution
Problem Statement
The density of mercury at 0°C is $13600$ kg/m³. Find its density at 100°C. ($\gamma_{Hg} = 1.8 \times 10^{-4}$ °C$^{-1}$)
Given Information
- $\rho_0 = 13600$ kg/m³ at 0°C
- $\gamma = 1.8 \times 10^{-4}$ °C$^{-1}$
- $\Delta T = 100°C$
Physical Concepts & Formulas
Mass is conserved; volume increases. If volume increases by factor $(1 + \gamma\Delta T)$, density decreases:
$$\rho = \frac{\rho_0}{1 + \gamma \Delta T}$$
Step-by-Step Solution
Step 1: Compute the volume expansion factor.
$$1 + \gamma \Delta T = 1 + 1.8 \times 10^{-4} \times 100 = 1 + 0.018 = 1.018$$
Step 2: Compute new density.
$$\rho = \frac{13600}{1.018}$$
Step 3: Calculate.
$$\rho = 13359 \text{ kg/m}^3 \approx 13360 \text{ kg/m}^3$$
Worked Calculation
$$\rho = \frac{13600}{1 + 1.8 \times 10^{-4} \times 100} = \frac{13600}{1.018} \approx 13360 \text{ kg/m}^3$$
Answer
$\rho \approx 13360$ kg/m³
Physical Interpretation
Mercury’s density drops by about 240 kg/m³ over a 100°C rise. This is why hydrometer readings (which measure liquid density) must be temperature-corrected. It also explains why mercury in thermometers expands visibly — the volumetric change is measurable and predictable.
Worked Calculation
Substituting all given numerical values with their units into the derived formula:
$$\text{Numerical result} = \text{given expression substituted with values}$$
Answer
$$\boxed{\rho = \frac{13600}{1 + 1.8 \times 10^{-4} \times 100} = \frac{13600}{1.018} \approx 13360 \text{ kg/m}^3}$$
Physical Interpretation
The answer should be checked for dimensional consistency and physical reasonableness: is the magnitude in the expected range for this type of problem? Does the answer change in the correct direction when parameters are varied (e.g., increasing mass should increase momentum, increasing distance should decrease field strength)? These sanity checks are as important as the calculation itself.
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