Thermal Properties of Matter

This chapter builds thermal physics as a tower, each floor resting on the one below. It starts by nailing down what temperature and heat actually mean, then shows how a thermometer converts a physical property into a number on a scale, and how the ideal-gas equation extends that idea to an absolute Kelvin scale with a true zero. With temperature defined, it asks what heat does: first it makes things bigger (thermal expansion of solids, liquids and gases, including water's odd behaviour near 4 degrees Celsius), then it studies how much heat is needed to change a body's temperature (specific and molar heat capacity, measured by calorimetry). Next comes what happens when heat is added at a fixed temperature — melting, boiling, sublimation and the latent heats that drive them, summarised in a temperature-versus-heat graph and a pressure-temperature phase diagram. The chapter closes with how heat travels from place to place — conduction, convection and radiation, including blackbody radiation, the Stefan-Boltzmann law and Wien's displacement law — and with Newton's law of cooling, which describes how fast any hot object settles down to room temperature.

Temperature, heat, and the scales that measure them

Temperature is a relative measure of hotness — a number assigned so that a hotter body reads higher than a colder one — while heat is the energy transferred between a system and its surroundings because of a temperature difference. A cup of hot tea cools because heat flows out of it into the cooler room; a glass of ice water warms because heat flows into it from the warmer room. Once the two are at the same temperature, this flow stops: that shared final temperature is called thermal equilibrium. The SI unit of heat is the joule (J), the same unit as any other form of energy, while temperature is measured in kelvin (K) or, informally, degrees Celsius. A thermometer works by picking some physical property that changes smoothly and reliably with temperature — the length of a mercury column, the resistance of a wire, the pressure of a gas at fixed volume — and calibrating it against two reproducible fixed points, historically the ice point (0 degrees Celsius) and the steam point (100 degrees Celsius) of water at standard pressure. Because the Fahrenheit scale places these same two points at 32 and 212 degrees, with 180 equal divisions between them against Celsius's 100, the two scales are related by a straight line:

where: $t_F$ is the temperature in degrees Fahrenheit and $t_C$ is the same temperature in degrees Celsius. Physically, this equation says both scales measure the identical physical hotness, just with different zero points and different step sizes — 100 Celsius degrees span exactly what 180 Fahrenheit degrees span.

Ideal gas equation, absolute temperature, and Kelvin

Liquid-in-glass thermometers disagree slightly away from their calibration points because different liquids expand differently, but a gas thermometer gives the same reading no matter which low-density gas fills it. All such gases combine Boyle's law ($PV = \text{constant}$ at fixed temperature) and Charles's law ($V/T = \text{constant}$ at fixed pressure) into a single ideal-gas equation,

where: $P$ is pressure, $V$ is volume, $\mu$ is the number of moles, $T$ is absolute temperature and $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ is the universal gas constant. Holding volume fixed, pressure grows linearly with temperature; extrapolating that straight line back to zero pressure — the pressure a gas would have if it kept obeying this law all the way down — defines a single, gas-independent minimum temperature: $-273.15\ ^\circ\text{C}$, called absolute zero. This is the zero of the Kelvin scale, related to Celsius by $T = t_C + 273.15$, with the same size of degree as Celsius, so a temperature difference of 1 K equals a temperature difference of 1 degree Celsius.

Thermal expansion

Almost every substance expands when heated because its molecules, on average, vibrate through a larger volume. For a rod of length $l$ and small temperature change $\Delta T$, the fractional change in length is directly proportional to $\Delta T$:

where: $\alpha_l$ is the coefficient of linear expansion (units $\text{K}^{-1}$), characteristic of the material — typical values are $2.5\times10^{-5}\ \text{K}^{-1}$ for aluminium down to $0.29\times10^{-5}\ \text{K}^{-1}$ for lead, with pyrex glass ($0.32\times10^{-5}\ \text{K}^{-1}$) prized for being especially low. Metals generally expand far more than glass for the same temperature rise, which is exactly why a metal lid on a glass jar loosens when run under hot water: the metal's larger $\alpha_l$ makes it expand faster than the glass beneath it. If the ends of an expanding rod are rigidly clamped, the prevented expansion shows up instead as compressive thermal stress, $\Delta F/A = Y\,\alpha_l\,\Delta T$, which can reach tens of megapascals for even a modest ten-degree rise in a steel rail. Liquids and gases show volume expansion only, since they take the shape of their container. Water is the chapter's famous anomaly: cooling from room temperature, its volume actually decreases down to 4 degrees Celsius (density maximum) before increasing again as it cools further toward freezing. This is why a lake freezes at the surface first — surface water cooling past 4 degrees Celsius becomes less dense and stays on top, insulating the water and life below, instead of sinking and freezing the lake from the bottom up. For an ideal gas at constant pressure, the volume expansivity is not a fixed material constant but depends on temperature itself, falling off as $\alpha_V = 1/T$ — at 0 degrees Celsius this gives $\alpha_V \approx 3.7\times10^{-3}\ \text{K}^{-1}$, roughly a hundred times larger than for a typical liquid or solid.

Specific heat capacity and calorimetry

Not every substance needs the same amount of heat to raise its temperature by the same amount: heating equal masses of water and mustard oil by the same 20 degrees Celsius takes visibly different times on the same burner. This "unique fingerprint" of a substance is its specific heat capacity,

where: $m$ is mass, $\mu$ is the number of moles, $\Delta Q$ is the heat absorbed or released, and $\Delta T$ is the resulting temperature change. $s$ (SI unit $\text{J kg}^{-1}\text{K}^{-1}$) is the specific heat capacity per unit mass, while $C$ (SI unit $\text{J mol}^{-1}\text{K}^{-1}$) is the molar specific heat capacity; for gases, $C$ must further be specified at constant pressure ($C_p$) or constant volume ($C_v$), since heating a gas that is free to expand takes more heat than heating the same gas held at fixed volume. Water's specific heat capacity, $4186\ \text{J kg}^{-1}\text{K}^{-1}$, is unusually high compared to metals or even ice, which is why water is used as a coolant and a heat reservoir, why coastal regions have milder climates than deserts, and why land heats and cools far faster than the sea. Calorimetry — literally "heat measurement" — puts this to work: when a hot body is placed in contact with a cooler one inside an insulated calorimeter, no heat escapes to the surroundings, so heat lost by the hot body exactly equals heat gained by everything else (the cooler substance and the calorimeter vessel itself, since it too warms up). This single bookkeeping principle, "heat lost equals heat gained," is the working tool used to measure an unknown specific heat capacity from a measured equilibrium temperature.

Change of state and latent heat

Heating a solid does not always raise its temperature. At a substance's melting point, added heat instead goes entirely into breaking the rigid bonds that hold it solid, and the temperature holds absolutely constant until every last bit has melted — solid and liquid coexist in thermal equilibrium throughout. The same happens at the boiling point during vaporisation, and, for substances like dry ice or iodine that skip the liquid phase entirely, during sublimation from solid straight to vapour. The amount of heat needed per unit mass to drive this change, without any temperature change, is the latent heat,

where: $m$ is the mass undergoing the change of state and $L$ is the latent heat (SI unit $\text{J kg}^{-1}$), split into the latent heat of fusion $L_f$ (solid to liquid) and latent heat of vaporisation $L_v$ (liquid to gas). For water at standard pressure, $L_f = 3.33\times10^5\ \text{J kg}^{-1}$ and $L_v = 22.6\times10^5\ \text{J kg}^{-1}$ — nearly seven times more heat is needed to boil a kilogram of water than to melt the same mass of ice, which is exactly why steam at 100 degrees Celsius carries so much more energy than boiling water at the same temperature, and why a steam burn is far more damaging. Both melting and boiling points shift with pressure. Increasing pressure raises the boiling point (the principle behind a pressure cooker) but lowers the melting point of ice slightly, which is what lets a wire loaded with weights slowly cut through an ice slab: the ice melts under the locally higher pressure beneath the wire and refreezes above it, a phenomenon called regelation. A pressure-temperature phase diagram maps out exactly which combinations of $P$ and $T$ correspond to solid, liquid or vapour, with the sublimation, fusion and vaporisation curves meeting at a single unique triple point — for water, $273.16\ \text{K}$ and $6.11\times10^{-3}\ \text{Pa}$ — where all three phases coexist simultaneously.

Heat transfer: conduction, convection and radiation

Heat moves from a hotter region to a cooler one by three distinct mechanisms. Conduction transfers heat through molecular collisions within a material with no bulk motion of matter — this is how the far end of a metal rod heats up when one end sits in a flame. In the steady state, with a bar of length $L$ and cross-sectional area $A$ held between two reservoirs at temperatures $T_C$ and $T_D$ ($T_C > T_D$),

where: $H$ is the heat current (rate of heat flow, in watts) and $K$ is the thermal conductivity of the material (SI unit $\text{W m}^{-1}\text{K}^{-1}$), a large value for metals like silver ($406$) and copper ($385$), and a small one for insulators like glass wool ($0.04$) or air ($0.024$). Convection, by contrast, transfers heat through the actual bulk motion of a fluid — hot fluid, being less dense, rises and is replaced by cooler, denser fluid sinking in, setting up a convection cycle; this drives sea breezes, trade winds and forced-air heating systems. Radiation needs no medium at all: every body emits electromagnetic radiation by virtue of its temperature (thermal radiation), which is how the Sun's energy crosses empty space to reach Earth. The full spectrum radiated by a body at temperature $T$ peaks at a wavelength given by Wien's displacement law, $\lambda_m T = \text{constant} = 2.9\times10^{-3}\ \text{m K}$ — hotter bodies glow at shorter, bluer wavelengths, which is why heated iron progresses from dull red through yellow to white as it gets hotter. The total power radiated obeys the Stefan-Boltzmann law,

where: $A$ is the surface area, $T$ is the absolute temperature, $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}$ is the Stefan-Boltzmann constant, and $e$ is the emissivity, a dimensionless number between 0 and 1 describing how close a real surface comes to a perfect radiator ($e=1$). A body in surroundings at temperature $T_s$ radiates and absorbs simultaneously, so its net rate of heat loss is $H = e\sigma A(T^4 - T_s^4)$ — this is how a human body, skin at about $301\ \text{K}$ in a $295\ \text{K}$ room with emissivity $0.97$ and surface area $1.9\ \text{m}^2$, is found to radiate away roughly $66\ \text{W}$, over half its resting metabolic output, which is exactly why arctic clothing adds a shiny reflective inner layer to cut radiative loss.

Newton's law of cooling

A hot cup of tea left on a table cools fast at first and then more and more slowly as it approaches room temperature — the rate of cooling itself depends on how far above the surroundings the body still is. Newton's law of cooling states this as a proportionality: the rate of loss of heat is directly proportional to the excess temperature over the surroundings, valid only for small temperature differences,

where: $T_2$ is the temperature of the cooling body, $T_1$ is the (constant) temperature of the surroundings, and $k$ is a positive constant depending on the body's surface area and nature. Because the rate of temperature fall is fastest when the excess temperature is largest, a plot of $\ln(T_2 - T_1)$ against time gives a straight line of negative slope — the experimental signature of this law, and the reason it is written and verified in logarithmic form.

Board level - Fitting an iron ring onto a wooden cart wheel

A blacksmith must fit an iron ring of diameter 5.231 m onto the 5.243 m rim of a wooden cart wheel at 27 degrees Celsius. To what temperature must the ring be heated so that it just slips over the rim? Take the coefficient of linear expansion of iron as $\alpha_l = 1.20\times10^{-5}\ \text{K}^{-1}$.

Step 1. The ring must expand from its initial diameter $L_{T_1} = 5.231\ \text{m}$ to the rim's diameter $L_{T_2} = 5.243\ \text{m}$, so apply the linear expansion relation in the form $L_{T_2} = L_{T_1}\left[1 + \alpha_l(T_2 - T_1)\right]$, with $T_1 = 27\ ^\circ\text{C}$.

Step 2. Substitute the numbers: $5.243 = 5.231\left[1 + 1.20\times10^{-5}(T_2 - 27)\right]$. Dividing both sides by $5.231$ gives $1.002293 = 1 + 1.20\times10^{-5}(T_2 - 27)$.

Step 3. Isolate the temperature rise: $T_2 - 27 = \dfrac{0.002293}{1.20\times10^{-5}} \approx 191\ ^\circ\text{C}$, so $T_2 \approx 218\ ^\circ\text{C}$. The ring must be heated to about 218 degrees Celsius before it will slip onto the rim.

Board level - Specific heat capacity of aluminium by calorimetry

A 0.047 kg aluminium sphere at 100 degrees Celsius is dropped into a 0.14 kg copper calorimeter containing 0.25 kg of water at 20 degrees Celsius. The mixture settles at 23 degrees Celsius. Given the specific heat capacities of water as $4186\ \text{J kg}^{-1}\text{K}^{-1}$ and of copper as $386\ \text{J kg}^{-1}\text{K}^{-1}$, find the specific heat capacity of aluminium.

Step 1. Heat lost by the aluminium sphere as it cools from 100 to 23 degrees Celsius is $Q_{\text{lost}} = m_{Al}\,s_{Al}\,\Delta T = 0.047 \times s_{Al} \times 77$.

Step 2. Heat gained by the water and the calorimeter as both warm from 20 to 23 degrees Celsius is $Q_{\text{gained}} = (m_w s_w + m_{Cu} s_{Cu})\,\Delta T_2 = (0.25\times4186 + 0.14\times386)\times 3$.

Step 3. Evaluate the gained side: $0.25\times4186 = 1046.5$ and $0.14\times386 = 54.0$, so $Q_{\text{gained}} = (1046.5+54.0)\times 3 = 3301.6\ \text{J}$. Since heat lost equals heat gained, $0.047\times s_{Al}\times77 = 3301.6$, giving $s_{Al} = \dfrac{3301.6}{3.619} \approx 912\ \text{J kg}^{-1}\text{K}^{-1}$. The specific heat capacity of aluminium works out to about $0.91\times10^3\ \text{J kg}^{-1}\text{K}^{-1}$, matching the standard tabulated value.

Competitive level - Heat to convert ice at minus 12 degrees Celsius fully to steam

3 kg of ice at $-12\ ^\circ\text{C}$ must be converted entirely to steam at $100\ ^\circ\text{C}$ at atmospheric pressure. Given $s_{\text{ice}} = 2100\ \text{J kg}^{-1}\text{K}^{-1}$, $s_{\text{water}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}$, $L_f = 3.35\times10^5\ \text{J kg}^{-1}$, and $L_{\text{steam}} = 2.256\times10^6\ \text{J kg}^{-1}$, find the total heat required.

Step 1. Warm the ice from $-12\ ^\circ\text{C}$ to $0\ ^\circ\text{C}$: $Q_1 = m\,s_{\text{ice}}\,\Delta T = 3\times2100\times12 = 75{,}600\ \text{J}$. Recognise immediately that this is a four-stage problem — warm, melt, warm, boil — and that each stage needs its own formula; mixing them up (for instance using the latent heat formula with a temperature difference) is the single biggest source of error here.

Step 2. Melt the ice at $0\ ^\circ\text{C}$: $Q_2 = m\,L_f = 3\times3.35\times10^5 = 1{,}005{,}000\ \text{J}$.

Step 3. Warm the resulting water from $0\ ^\circ\text{C}$ to $100\ ^\circ\text{C}$: $Q_3 = m\,s_{\text{water}}\,\Delta T = 3\times4186\times100 = 1{,}255{,}800\ \text{J}$; then boil it away at $100\ ^\circ\text{C}$: $Q_4 = m\,L_{\text{steam}} = 3\times2.256\times10^6 = 6{,}768{,}000\ \text{J}$.

Step 4. Add all four stages: $Q = Q_1+Q_2+Q_3+Q_4 = 75{,}600 + 1{,}005{,}000 + 1{,}255{,}800 + 6{,}768{,}000 \approx 9.1\times10^6\ \text{J}$. The total heat required is about $9.1\times10^6\ \text{J}$, or 9.1 MJ. Timing note: about two minutes. Sketch the temperature-versus-heat staircase mentally first — warm, plateau, warm, plateau — before touching a calculator, so you never mix a specific-heat term with a latent-heat term. Notice also that $Q_4$ (boiling) dwarfs the other three combined: nearly three-quarters of the total energy goes into vaporising the water, not warming anything, a fact worth having memorised for quick sanity checks.

Competitive level - Newton's law of cooling by the ratio method

A pan of hot food cools from 94 degrees Celsius to 86 degrees Celsius in 2 minutes, with the room at 20 degrees Celsius. How long will the same pan take to cool from 71 degrees Celsius to 69 degrees Celsius?

Step 1. Newton's law of cooling is only accurate for the average temperature during a small interval, so approximate the rate as $\dfrac{\Delta T}{\text{time}} = k\,(T_{\text{avg}} - T_1)$. For the first interval, the average temperature is $\dfrac{94+86}{2} = 90\ ^\circ\text{C}$, which is $70\ ^\circ\text{C}$ above the room's 20 degrees Celsius, and the temperature fell by $8\ ^\circ\text{C}$ in 2 minutes.

Step 2. For the second interval, the average temperature is $\dfrac{71+69}{2} = 70\ ^\circ\text{C}$, which is $50\ ^\circ\text{C}$ above room temperature, with the temperature falling by $2\ ^\circ\text{C}$ in an unknown time $t$.

Step 3. Write both intervals using the same constant $k$: $\dfrac{8}{2} = k(70)$ and $\dfrac{2}{t} = k(50)$. Dividing the second equation by the first eliminates $k$ entirely: $\dfrac{2/t}{8/2} = \dfrac{50}{70}$, so $\dfrac{2/t}{4} = \dfrac{5}{7}$, giving $\dfrac{2}{t} = \dfrac{20}{7}$ and $t = \dfrac{14}{20} = 0.7\ \text{min}$. The pan takes about 0.7 minutes, or 42 seconds, to cool from 71 to 69 degrees Celsius. Timing note: well under a minute once you recognise the ratio trick. The entire point of this method is to never actually solve for $k$ — set up rate equals $k$ times excess temperature for both intervals, using the average temperature of each interval (not the starting or ending value), and divide one equation by the other so $k$ cancels. Trying to integrate the exponential form here would cost far more time than the mark is worth.

Students lose marks here in a few specific, repeatable ways. The most frequent is treating heat and temperature as interchangeable — saying a body "contains" heat, when heat is energy in transit and temperature is only what a thermometer reads. A close second is forgetting that a phase change carries its own separate latent-heat term: in a multi-stage problem (ice below zero warming, melting, warming again, possibly boiling), every plateau needs its own $Q = mL$ alongside the sloped segments' $Q = ms\Delta T$, and mixing the two formulas is the single most common numerical error. Sign errors plague calorimetry too: heat lost by the hotter object must equal heat gained by everything cooler, including the calorimeter vessel itself, which is very often forgotten. In thermal expansion, students frequently write $\alpha_V = \alpha_l$ instead of $3\alpha_l$, or use $\alpha_l$ for an area problem instead of $2\alpha_l$. In conduction, forgetting that heat current $H$, not temperature, stays constant along a composite bar leads to the wrong junction condition. Finally, Newton's law of cooling gets quoted as universally true, when it is really a small-excess-temperature approximation, and students forget to use each interval's average temperature rather than its starting value.

• Convert confidently between Celsius, Fahrenheit and Kelvin, and explain why absolute zero is defined by extrapolating a gas thermometer's pressure to zero.
• State the ideal-gas equation $PV = \mu RT$ and explain why a gas thermometer gives the same reading regardless of which gas it uses.
• Compute linear, area and volume thermal expansion for a solid, and know the relations $\Delta A/A = 2\alpha_l\Delta T$ and $\alpha_V = 3\alpha_l$.
• Explain water's anomalous expansion between 0 and 4 degrees Celsius and why it lets lakes freeze from the top down.
• Calculate an unknown specific or molar heat capacity from a calorimetry experiment using heat lost equals heat gained.
• Track heat through a multi-stage problem that mixes temperature changes and phase changes, using $Q = ms\Delta T$ and $Q = mL$ correctly at each stage.
• Compute steady-state conduction through a single bar and through bars in series, including the junction temperature and effective conductivity.
• Apply the Stefan-Boltzmann law and Wien's displacement law to a radiating body, and distinguish conduction, convection and radiation by whether they need a medium.
• Apply Newton's law of cooling, including the ratio method for eliminating the unknown cooling constant between two time intervals.

Ask the tutor: Want to work through a mixed calorimetry-and-conduction problem step by step, or see how the phase diagram of water connects to what you just learned about melting, boiling and the triple point?