Temperature and its measurement

Heat vs. temperature. Heat is energy in transit, flowing spontaneously from a hotter body to a colder one because of a temperature difference; it is measured in joules. Temperature is not energy at all — it is the state variable that determines the direction heat will flow, and (as formalised later by the zeroth law) it is the property that decides whether two bodies are in thermal equilibrium. A large tub of lukewarm water can hold far more heat energy than a small red-hot needle, even though the needle is at the higher temperature — heat depends on mass and heat capacity too, temperature does not.

Building a scale from fixed points. Every practical scale is defined the same way: pick two reproducible reference temperatures (the fixed points) and divide the interval between them into a chosen number of equal degrees.

• Celsius ($^\circ\text{C}$): ice point $= 0^\circ\text{C}$, steam point $= 100^\circ\text{C}$, split into 100 equal divisions.
• Fahrenheit ($^\circ\text{F}$): ice point $= 32^\circ\text{F}$, steam point $= 212^\circ\text{F}$, split into 180 equal divisions.
• Kelvin (K), the SI base unit: its zero, absolute zero, is the extrapolated point at which an ideal gas would have zero pressure and volume — the lowest temperature theoretically possible, where molecular kinetic energy is minimal. Kelvin has no negative values and carries no degree symbol.

Because all three scales are just different rulers laid over the same physical interval, the fraction of the way from one fixed point to the other must agree across scales:

where: $C$, $F$, $K$ are the same temperature read on the Celsius, Fahrenheit and Kelvin scales respectively. This simplifies to the two forms used constantly in numericals:

(more precisely $K = C + 273.15$, but $+273$ is the standard ISC-level approximation).

Why the ideal gas equation matters here. A mercury thermometer's reading depends on how mercury specifically expands with temperature; an alcohol thermometer depends on alcohol's expansion; a resistance thermometer depends on how a particular wire's resistance changes. Different thermometric substances agree only at the two calibration fixed points — between and beyond them, their readings can drift apart slightly, because no two properties (expansion, resistance, pressure) vary with temperature in exactly the same way. The ideal gas equation

where: $p$ is pressure, $V$ is volume, $n$ is the number of moles, $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ is the universal gas constant, and $T$ is temperature in kelvin — solves this problem. A constant-volume gas thermometer measures pressure at fixed volume; extrapolated to zero pressure, it defines a temperature scale that agrees with the kelvin scale and pins absolute zero as the temperature at which an ideal gas's pressure would fall to zero. A dilute real gas behaves very nearly like an ideal gas, so this scale is almost independent of which gas is used — unlike two liquid thermometers filled with different liquids, which can disagree away from the fixed points.

A patient's body temperature reads $37^\circ\text{C}$ on a clinical thermometer. Express this in Fahrenheit and Kelvin, and then find the single temperature at which the Celsius and Fahrenheit scales give numerically identical readings.

Step 1. Convert to Fahrenheit using $\dfrac{C}{5} = \dfrac{F-32}{9}$ with $C = 37$: $F - 32 = \dfrac{37 \times 9}{5} = 66.6$, so $F = 98.6^\circ\text{F}$.

Step 2. Convert to Kelvin using $K = C + 273$: $K = 37 + 273 = 310\ \text{K}$.

Step 3. For the coincidence point, set $C = F$ in $\dfrac{C}{5} = \dfrac{F-32}{9}$: $9C = 5(C - 32) = 5C - 160$, so $4C = -160$ and $C = -40$.

Answer: $37^\circ\text{C} = 98.6^\circ\text{F} = 310\ \text{K}$, and the Celsius and Fahrenheit scales coincide at $-40^\circ$ (i.e. $-40^\circ\text{C} = -40^\circ\text{F}$).

Competitive stretch

A gas thermometer measures pressure $p_1 = 1.20\times10^5\ \text{Pa}$ at $T_1 = 300\ \text{K}$, at constant volume. At what pressure would it read $T_2 = 0\ \text{K}$ on the ideal-gas extrapolation? Since $pV = nRT$ with $V$, $n$ constant, $p \propto T$, so $p_2 = p_1 \times (T_2/T_1) = 1.20\times10^5 \times 0 = 0\ \text{Pa}$ — the pressure extrapolates to exactly zero at absolute zero, which is precisely how the ideal gas scale defines $0\ \text{K}$ in the first place.

• Writing kelvin as "$^\circ$K" or allowing a negative kelvin value — kelvin carries no degree symbol and cannot go below zero.
• Assuming a body with "more heat" is automatically "hotter" — heat depends on mass and heat capacity as well as temperature; a large lukewarm object can hold more heat energy than a small very hot one.
• Mixing up which interval (100, 180, or 100) belongs to which scale in the conversion formula — if unsure, re-derive from the fixed points (0-100 Celsius, 32-212 Fahrenheit) rather than guessing.
• Treating a real gas thermometer's reading as identical to the true ideal-gas (kelvin) scale — real gases only approximate ideal behaviour, and the approximation worsens at low temperature or high pressure.

Ask the tutor: if the fixed-point logic behind the conversion formula or the physical meaning of absolute zero from the ideal gas equation still feels shaky, ask the AI tutor to walk through another gas-thermometer example step by step.