1. Overview
Temperature is a macroscopic measure of the average kinetic energy of the particles within a system. It determines the direction of net thermal energy flow between two bodies in thermal contact; energy always flows from a region of higher temperature to a region of lower temperature. When two objects reach the same temperature, they are in thermal equilibrium, and the net flow of energy becomes zero. To measure temperature accurately and consistently, we rely on physical properties that change predictably with temperature (thermometric properties) and a standardized scale that is independent of the specific material used (the thermodynamic scale).
Key Definitions
- Thermal Equilibrium: A state in which two or more objects in thermal contact have the same temperature, resulting in no net flow of thermal energy between them.
- Thermometric Property: A physical property of a substance that varies continuously, uniquely, and measurably with changes in temperature.
- Absolute Zero: The theoretical lowest possible temperature ($0\text{ K}$ or $-273.15\text{ }^\circ\text{C}$), defined as the temperature at which a system possesses minimum internal energy.
- Thermodynamic (Absolute) Temperature Scale: A temperature scale that is independent of the properties of any specific substance and has a fixed lower limit of absolute zero.
- Triple Point of Water: The unique temperature and pressure at which water exists simultaneously as a solid, liquid, and gas in thermal equilibrium. It is defined as exactly $273.16\text{ K}$.
- Fixed Point: A standard, easily reproducible temperature (such as the boiling point of water at standard pressure) used to calibrate temperature scales.
Content
3.1 Physical Properties Used to Measure Temperature
A thermometer is constructed by choosing a substance with a thermometric property and calibrating it against known fixed points. The Cambridge 9702 syllabus identifies four specific properties:
| Property | Instrument | Physical Basis |
|---|---|---|
| Density of a liquid | Liquid-in-glass thermometer | As temperature increases, the liquid expands, its density decreases, and the volume increases, causing the liquid to rise up a narrow capillary tube. |
| Volume of a gas | Constant-pressure gas thermometer | For an ideal gas at constant pressure, the volume is directly proportional to the thermodynamic temperature ($V \propto T$). |
| Resistance of a metal | Resistance thermometer (e.g., Platinum) | As temperature increases, lattice ions vibrate more vigorously, increasing the scattering of conduction electrons, which increases the electrical resistance. |
| E.m.f. of a thermocouple | Thermocouple | Two dissimilar metals joined at two junctions create a potential difference (e.m.f.) when the junctions are at different temperatures. |
3.2 The Disagreement Between Thermometers
If we calibrate a liquid-in-glass thermometer and a resistance thermometer using the same two fixed points (e.g., the ice point $0\text{ }^\circ\text{C}$ and the steam point $100\text{ }^\circ\text{C}$), they will both agree at exactly $0\text{ }^\circ\text{C}$ and $100\text{ }^\circ\text{C}$.
However, at an intermediate temperature (e.g., $50\text{ }^\circ\text{C}$), they may give slightly different readings. This occurs because:
- Different thermometric properties do not necessarily vary linearly with each other.
- One property might change linearly with thermodynamic temperature while the other follows a more complex relationship (e.g., quadratic).
- Therefore, an "empirical" scale (one based on a specific substance) is always slightly arbitrary.
3.3 The Thermodynamic Temperature Scale
To overcome the limitations of substance-dependent scales, physicists use the Thermodynamic Scale (the Kelvin scale).
Key Characteristics:
- Independence: It does not depend on the property of any particular substance. It is derived from the laws of thermodynamics (specifically the efficiency of a reversible heat engine).
- Absolute Zero: It has a natural starting point ($0\text{ K}$) where the pressure of an ideal gas would theoretically be zero.
- The Fixed Point: It uses only one fixed point for its definition: the triple point of water, which is defined as $273.16\text{ K}$.
Why $273.16\text{ K}$ and not $273.15\text{ K}$? The triple point of water is $0.01\text{ }^\circ\text{C}$. Since $0\text{ }^\circ\text{C}$ is defined as $273.15\text{ K}$, the triple point must be $273.15 + 0.01 = 273.16\text{ K}$.
3.4 Conversion between Kelvin and Celsius
The Kelvin ($\text{K}$) and the degree Celsius ($^\circ\text{C}$) are related by a simple additive offset. Crucially, a change of $1\text{ K}$ is exactly equal to a change of $1\text{ }^\circ\text{C}$.
The Conversion Formula: $$T / \text{K} = \theta / ^\circ\text{C} + 273.15$$ (Note: This formula is not provided on the data sheet; it must be memorized.)
Where:
- $T$ = Thermodynamic temperature in Kelvin ($\text{K}$)
- $\theta$ = Temperature in degrees Celsius ($^\circ\text{C}$)
3.5 Absolute Zero
Absolute zero is the lower limit of the thermodynamic temperature scale.
- Numerical Value: $0\text{ K}$ or $-273.15\text{ }^\circ\text{C}$.
- Physical Significance: It is the temperature at which the atoms or molecules in a substance have minimum internal energy.
- Kinetic Theory Perspective: For an ideal gas, the average kinetic energy of the molecules is proportional to the thermodynamic temperature ($E_k = \frac{3}{2}kT$). At $0\text{ K}$, the translational kinetic energy of the molecules would be zero.
- Pressure Perspective: Since $P = \frac{1}{3}\rho \langle c^2 \rangle$, if the molecular speed $c$ is zero, the pressure exerted by the gas is zero.
4. Worked Examples
Worked example 1 — Basic Unit Conversion
A sample of liquid oxygen is stored at a temperature of $-183.00\text{ }^\circ\text{C}$. Calculate this temperature in Kelvin.
Step 1: State the conversion equation $$T / \text{K} = \theta / ^\circ\text{C} + 273.15$$
Step 2: Substitute the Celsius value $$T = -183.00 + 273.15$$
Step 3: Calculate the final value $$T = 90.15\text{ K}$$
Worked example 2 — Thermocouple Linearity
A thermocouple is used to measure the temperature of a furnace. The e.m.f. produced is $0.0\text{ mV}$ at $0\text{ }^\circ\text{C}$ and $24.6\text{ mV}$ at $100\text{ }^\circ\text{C}$. When placed in the furnace, the e.m.f. is $18.5\text{ mV}$. Calculate the temperature of the furnace, assuming the e.m.f. varies linearly with temperature.
Step 1: Use the linear interpolation formula for an empirical scale $$\theta = \frac{E_\theta - E_0}{E_{100} - E_0} \times 100$$
Step 2: Substitute the given e.m.f. values $$\theta = \frac{18.5 - 0.0}{24.6 - 0.0} \times 100$$
Step 3: Perform the calculation $$\theta = \frac{18.5}{24.6} \times 100 = 75.203...$$
Step 4: Final answer to appropriate significant figures $$\theta = 75.2\text{ }^\circ\text{C}$$
Worked example 3 — Resistance Thermometer
The resistance $R_\theta$ of a platinum wire at temperature $\theta$ (in $^\circ\text{C}$) is given by $R_\theta = R_0(1 + \alpha\theta)$, where $\alpha = 3.9 \times 10^{-3}\text{ }^\circ\text{C}^{-1}$. If the resistance is $100.0\text{ }\Omega$ at $0\text{ }^\circ\text{C}$, calculate the temperature when the resistance is $145.0\text{ }\Omega$.
Step 1: Rearrange the equation to solve for $\theta$ $$R_\theta = R_0 + R_0\alpha\theta$$ $$\theta = \frac{R_\theta - R_0}{R_0\alpha}$$
Step 2: Substitute the values $$\theta = \frac{145.0 - 100.0}{100.0 \times (3.9 \times 10^{-3})}$$
Step 3: Calculate $$\theta = \frac{45.0}{0.39} = 115.38...$$
Step 4: Final answer $$\theta = 115\text{ }^\circ\text{C}$$
Worked example 4 — Absolute Zero Extrapolation
In an experiment with a constant-volume gas thermometer, the pressure of a gas is $1.20 \times 10^5\text{ Pa}$ at $100\text{ }^\circ\text{C}$ and $0.88 \times 10^5\text{ Pa}$ at $0\text{ }^\circ\text{C}$. Use these values to estimate the value of absolute zero in degrees Celsius.
Step 1: Understand the method Absolute zero is the temperature where pressure $P = 0$. We can find the gradient of the $P$-$\theta$ graph and extrapolate.
Step 2: Calculate the gradient ($m$) of the $P$ vs $\theta$ line $$m = \frac{\Delta P}{\Delta \theta} = \frac{1.20 \times 10^5 - 0.88 \times 10^5}{100 - 0}$$ $$m = \frac{0.32 \times 10^5}{100} = 320\text{ Pa }^\circ\text{C}^{-1}$$
Step 3: Use the equation of the line ($P = m\theta + c$) At $0\text{ }^\circ\text{C}$, $P = 0.88 \times 10^5$, so $c = 0.88 \times 10^5$. $$P = 320\theta + 88000$$
Step 4: Set $P = 0$ and solve for $\theta$ $$0 = 320\theta + 88000$$ $$\theta = -\frac{88000}{320} = -275\text{ }^\circ\text{C}$$
Answer: The estimated absolute zero is $-275\text{ }^\circ\text{C}$.
Key Equations
| Equation | Description | Status |
|---|---|---|
| $T / \text{K} = \theta / ^\circ\text{C} + 273.15$ | Conversion between Kelvin and Celsius. | Memorize |
| $\Delta T = \Delta \theta$ | A change in Kelvin is equal to a change in Celsius. | Understand |
| $\theta = \frac{X_\theta - X_0}{X_{100} - X_0} \times 100$ | Calculating temperature on a linear empirical scale. | Memorize |
| $PV = nRT$ | Ideal Gas Law (requires $T$ in Kelvin). | Data Sheet |
Common Mistakes to Avoid
- ❌ Wrong: Using the degree symbol for Kelvin (e.g., $300\text{ }^\circ\text{K}$).
- ✅ Right: Kelvin is an absolute unit; write it as $300\text{ K}$.
- ❌ Wrong: Using $273$ for conversions when the question provides data to two decimal places.
- ✅ Right: Use $273.15$ for precision in A-Level calculations.
- ❌ Wrong: Stating that particles have "zero energy" at absolute zero.
- ✅ Right: Particles have minimum internal energy. (Quantum mechanics dictates they still have "zero-point energy").
- ❌ Wrong: Assuming all thermometers agree at all temperatures.
- ✅ Right: They only agree at the fixed points used for calibration.
- ❌ Wrong: Forgetting to convert Celsius to Kelvin when using the Ideal Gas Law ($PV=nRT$).
- ✅ Right: Always check units before substituting into thermal physics formulas.
Exam Tips
- Defining the Thermodynamic Scale: If asked why the Kelvin scale is used, the most important marking point is that it is independent of the properties of any substance.
- Thermal Equilibrium Keywords: When defining thermal equilibrium, you must state there is no net flow of thermal energy. Simply saying "no energy flows" is technically incorrect, as energy is still exchanged at the molecular level.
- Significant Figures in Conversions: When adding $273.15$ to a Celsius temperature, follow the rules for addition: the result should have the same number of decimal places as the least precise measurement.
- Example: $25\text{ }^\circ\text{C} + 273.15 = 298\text{ K}$ (rounded to nearest whole number).
- Example: $25.00\text{ }^\circ\text{C} + 273.15 = 298.15\text{ K}$.
- Thermometric Properties: Be specific. If asked for a property, do not just name the device.
- Instead of "Thermocouple", write "e.m.f. of a thermocouple".
- Instead of "Resistance thermometer", write "resistance of a metal wire".
- Graphing Absolute Zero: In Paper 5 or structured questions, you may be asked to extrapolate a graph to find absolute zero. Ensure your line of best fit is drawn with a ruler and extends clearly to the x-axis (temperature) where the y-axis (pressure or volume) is zero.