Thermal Properties of Matter · Lesson 1 of 8
Temperature, Thermometers and the Absolute Scale
“Distinguish heat from temperature and build temperature scales using thermometric properties, ideal gases and fixed-point calibration.”
• Explain the direction of heat transfer and distinguish temperature from transferred energy. • Derive conversions between Celsius, Fahrenheit and Kelvin, including temperature intervals. • Use Boyle’s law, Charles’ law and the ideal-gas equation with absolute temperature. • Calibrate resistance and gas thermometers and explain the role of triple points and low-density extrapolation.
Heat and temperature answer different questions
A cup of hot tea cools on a table, while ice-cold water warms in the same room. In both cases the direction of energy transfer is determined by a temperature difference. Temperature describes the thermal condition that tells which body is hotter. Heat is energy transferred between systems because of that difference. Its SI unit is the joule; the SI unit of thermodynamic temperature is the kelvin, symbol K.
Heat is not a material stored inside an object, nor is temperature an amount of energy. A small hot object can transfer less energy on cooling than a large moderately warm one, depending on mass, material and final conditions. The amount transferred and the temperature change are related through a material’s heat capacity, which we will develop later. Heat input may raise temperature, expand a substance or change its state; a temperature rise is not inevitable.
Two bodies in thermal contact approach a common temperature when their surroundings do not continually force them apart. At thermal equilibrium there is no net heat transfer caused by a temperature difference. Touch provides only a limited, sometimes misleading temperature sense: the rate at which a surface transfers energy to or from the skin also matters. Scientific measurement therefore uses a reproducible physical property rather than how hot or cold a sample happens to feel.
A thermometer converts a changing property into a number
A thermometric property changes predictably with temperature. A liquid-in-glass thermometer uses the volume of a liquid, observed as the length of its column in a narrow tube. Electrical resistance and the pressure of a fixed amount of gas at fixed volume are other examples. The instrument must reach thermal equilibrium with the sample, and its calibration must be valid in the range being measured.
To construct a simple linear scale, choose two reproducible reference conditions and assign temperatures to them. Those choices fix the numerical origin and the size of the interval. The original Celsius scale used the freezing and boiling conditions of pure water under standard atmospheric pressure, labelled 0 °C and 100 °C. The corresponding Fahrenheit labels are 32 °F and 212 °F. The same physical interval is therefore divided into 100 Celsius intervals or 180 Fahrenheit intervals.
Subtract the conversion equation for two temperatures and the offset cancels. A rise of 10 °C is a rise of 18 °F, not 50 °F. In contrast, the reading 10 °C corresponds to 50 °F. Keep a temperature reading separate from a temperature difference. Liquid thermometers that agree at the two reference points can still differ between or beyond them because real liquids do not expand in exactly the same way at every temperature.
Gas laws motivate an absolute temperature scale
For a fixed amount of sufficiently dilute gas, Boyle’s law states that PV is constant when temperature is fixed. At fixed pressure, Charles’ law states that V/T is constant when T is an absolute temperature. These controlled experimental relations combine into PV/T = constant for that fixed sample. Doubling the amount of ideal gas doubles this constant, giving the general ideal-gas equation.
The gas laws concern specified conditions, not simultaneous claims that P, V and T are all fixed. If V and n are held fixed, pressure is proportional to T. If P and n are fixed, volume is proportional to T. Celsius readings cannot be used in these proportional ratios: a gas at 20 °C does not have twice the absolute temperature of one at 10 °C.
A plot of constant-volume gas pressure against Celsius temperature is approximately straight over a suitable range. Extending the idealized line back gives zero pressure at −273.15 °C. Different dilute gases lead to the same limiting zero. This motivates the absolute scale: set that extrapolated zero to 0 K and give a kelvin the same interval size as a Celsius degree. Real gases deviate and can condense before that extrapolated point, so this is not an instruction to follow an actual gaseous sample continuously down to zero pressure.
| Condition | Kelvin / K | Celsius / °C | Fahrenheit / °F |
|---|---|---|---|
| Absolute zero | 0 | −273.15 | −459.67 |
| Approximate normal ice point | 273.15 | 0 | 32 |
| Approximate normal steam point | 373.15 | 100 | 212 |
Problem
The triple-point temperatures of neon and carbon dioxide are given as 24.57 K and 216.55 K. Express both on Celsius and Fahrenheit scales.
- 1.Neon: tC = 24.57 − 273.15 = −248.58 °C. Then tF = (9/5)(−248.58) + 32 = −415.44 °F.
- 2.Carbon dioxide: tC = 216.55 − 273.15 = −56.60 °C. Then tF = (9/5)(−56.60) + 32 = −69.88 °F.
- 3.Negative Celsius or Fahrenheit readings are compatible with positive absolute temperatures. Keep the sign through the full calculation.
Problem
Absolute scales A and B assign water’s triple point readings 200 A and 350 B. Find the relation between readings TA and TB. Also find water’s triple-point reading on an absolute scale with Fahrenheit-sized intervals.
- 1.Both absolute scales start at the same physical zero. Their readings at any temperature therefore have a constant ratio: TA/TB = 200/350 = 4/7.
- 2.Thus TA = (4/7)TB. There is no additional offset because the zeros coincide.
- 3.A Fahrenheit-sized interval is 5/9 of a kelvin, so the corresponding absolute reading is (9/5)T. At the conventional triple-point value 273.16 K, this gives 491.688, about 491.69.
- 4.Do not add 32: that offset belongs to the ordinary Fahrenheit scale, whose zero is not absolute zero.
Fixed points: textbook convention and present SI
The triple point is the condition at which solid, liquid and vapour of a specified substance coexist in equilibrium. It fixes both temperature and pressure for that substance. It is more reproducible for thermometry than an unspecified melting or boiling condition, since melting and boiling temperatures depend on the pressure imposed. Water’s familiar triple-point temperature is 273.16 K, or 0.01 °C. The 273.15 offset in the Celsius relation makes the Celsius zero differ from this triple point by 0.01 degree; replacing the offset by 273.16 would change the scale.
The chapter’s fixed-point discussion reflects the older definition of the kelvin through water’s triple point. Since 2019 the SI kelvin is defined by fixing the Boltzmann constant at 1.380649 × 10⁻²³ J/K. The practical ITS-90 scale still assigns 273.16 K to the water triple point. In the textbook’s absolute-scale construction, absolute zero supplies the conceptual zero anchor; it is not another readily realizable laboratory fixed point. The chapter’s numerical exercises can still use their stated 273.16 K reference.
Calibration without knowing every instrument constant
If an instrument property X varies linearly with temperature, eliminate unknown constants by comparing fractional changes between two calibration points. Equal fractional positions between the two X readings correspond to equal fractional positions between the two temperatures. This is an interpolation rule, not a guarantee that every thermometer remains linear outside its calibration range.
Problem
A thermometer reads 101.6 Ω at 273.16 K and 165.5 Ω at 600.5 K. Its resistance at an unknown temperature is 123.4 Ω. Use the stated linear approximation.
- 1.The fraction of the resistance interval is (123.4 − 101.6)/(165.5 − 101.6) = 21.8/63.9.
- 2.The corresponding temperature is T = 273.16 + (21.8/63.9)(600.5 − 273.16) = 384.83 K.
- 3.The result lies between the two calibration temperatures, as its resistance lies between the two calibration resistances.
Problem
At water’s reference triple point, thermometer A reads 1.250 × 10⁵ Pa and B reads 0.200 × 10⁵ Pa. At the sulphur melting condition they read 1.797 × 10⁵ Pa and 0.287 × 10⁵ Pa. Find their inferred temperatures.
- 1.For A, TA = 273.16(1.797/1.250) = 392.695 K.
- 2.For B, TB = 273.16(0.287/0.200) = 391.985 K.
- 3.Perfect ideal gases would agree. A small discrepancy in the observed instruments reflects nonideal gas behaviour at finite density, rather than necessarily a defective thermometer.
- 4.Repeat with successively smaller reference pressures/densities and extrapolate the inferred readings to the zero-density limit, where gas-specific departures become negligible.
Quick Quiz
Quiz
Which statement defines heat?
Heat names a mode of energy transfer caused by temperature difference, not a material or a state-variable reading.
A temperature rises by 15 °C. What is the same rise in Fahrenheit intervals?
For a difference, the offset cancels: ΔtF = (9/5) × 15 = 27 °F.
A fixed amount of ideal gas is heated at constant volume from 300 K to 450 K. Its absolute pressure becomes
At fixed n and V, P/T is constant, so P2/P1 = 450/300.
Why does extrapolating a gas-thermometer line not prove that a real gas stays gaseous down to absolute zero?
The ideal extrapolation identifies a limiting scale; real phase changes and nonideality intervene.
Why is water’s triple point 0.01 °C rather than 0 °C when assigned the conventional reading 273.16 K?
tC = T − 273.15 gives 0.01 °C at 273.16 K.
Practice Problems
- Convert 310.15 K to Celsius and Fahrenheit. Separately convert a temperature rise of 12 K to Celsius and Fahrenheit intervals, explaining why the offsets are not used for intervals.
- Two absolute temperature scales label the same reference condition 300 X and 450 Y. Derive their general conversion and determine the X reading corresponding to 600 Y.
- A linear resistance thermometer reads 80.0 Ω at 280 K and 120.0 Ω at 380 K. Find the temperature when it reads 98.0 Ω and state the linearity assumption used.
- A constant-volume dilute-gas thermometer has pressure 1.00 × 10⁵ Pa at 273.16 K and 1.35 × 10⁵ Pa at an unknown condition. Calculate the inferred temperature. Explain why gauge pressure would not be appropriate in this ratio.
- Explain why two liquid thermometers may agree at their calibration points but differ elsewhere, while dilute-gas thermometers approach a common scale as density is reduced. Distinguish a reproducible practical fixed point from the present SI definition of the kelvin.
Key Takeaways
• Temperature determines thermal equilibrium and the direction of heat transfer; heat is transferred energy. • Scale offsets affect readings, while interval ratios determine temperature differences. • Use kelvins and absolute pressures in ideal-gas proportionalities. • Fixed-point calibration, ideal-gas extrapolation and the present Kelvin definition are related but distinct ideas.
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Thermal Expansion and Thermal Stress