Problem Statement
A glass flask is completely filled with mercury at 0°C. What fraction of mercury overflows when heated to 100°C? ($\gamma_{Hg} = 1.8 \times 10^{-4}$ °C$^{-1}$, $\gamma_{glass} = 2.7 \times 10^{-5}$ °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
A glass flask is completely filled with mercury at 0°C. What fraction of mercury overflows when heated to 100°C? ($\gamma_{Hg} = 1.8 \times 10^{-4}$ °C$^{-1}$, $\gamma_{glass} = 2.7 \times 10^{-5}$ °C$^{-1}$)
Given Information
- $\gamma_{Hg} = 1.8 \times 10^{-4}$ °C$^{-1}$
- $\gamma_{glass} = 2.7 \times 10^{-5}$ °C$^{-1}$ (the container also expands)
- $\Delta T = 100°C$
Physical Concepts & Formulas
Both the mercury and the flask expand. The apparent (effective) expansion coefficient of mercury relative to glass is:
$$\gamma_{apparent} = \gamma_{Hg} – \gamma_{glass}$$
Fraction overflow = $\gamma_{apparent} \times \Delta T$
Step-by-Step Solution
Step 1: Compute apparent expansion coefficient.
$$\gamma_{app} = 1.8 \times 10^{-4} – 2.7 \times 10^{-5} = 1.8 \times 10^{-4} – 0.27 \times 10^{-4} = 1.53 \times 10^{-4} \text{ °C}^{-1}$$
Step 2: Compute fractional overflow.
$$\frac{\Delta V}{V_0} = \gamma_{app} \times \Delta T = 1.53 \times 10^{-4} \times 100$$
Step 3: Result.
$$\frac{\Delta V}{V_0} = 1.53 \times 10^{-2} = 1.53\%$$
Worked Calculation
$$\gamma_{app} = (1.80 – 0.27) \times 10^{-4} = 1.53 \times 10^{-4} \text{ °C}^{-1}$$
$$\text{Fraction} = 1.53 \times 10^{-4} \times 100 = 0.0153 \approx 1.53\%$$
Answer
Fraction overflow $\approx 1.53\%$
Physical Interpretation
Mercury expands much more than glass, so the net overflow is significant. Thermometers exploit this: mercury rising in a glass capillary directly represents the difference in their expansion coefficients. The glass expands slightly too, but mercury expands ~6.7 times more, producing a clear visual reading.
Worked Calculation
Substituting all given numerical values with their units into the derived formula:
$$\text{Numerical result} = \text{given expression substituted with values}$$
Answer
$$\boxed{\text{Fraction} = 1.53 \times 10^{-4} \times 100 = 0.0153 \approx 1.53\%}$$
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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