Thermal Properties of Matter

A thermometer uses a property that changes steadily with temperature: the length of a mercury thread, the pressure of a gas held at fixed volume, or the resistance of a platinum wire. Two fixed points calibrate it. The lower fixed point (LFP) is the ice point and the upper fixed point (UFP) is the steam point, both at 1 atm.

If the property changes linearly, a reading that sits the same fraction of the way from LFP to UFP means the same temperature on every scale:

\[ \frac{\text{reading}-\text{LFP}}{\text{UFP}-\text{LFP}}\ \text{is the same on every scale.} \]

So \(\frac{C}{100}=\frac{F-32}{180}=\frac{K-273.15}{100}=\frac{R}{80}\). The same rule converts readings of a faulty or home-made thermometer: use its own marks for LFP and UFP.

• Celsius and Fahrenheit agree at −40°. Put \(F=C\) in \(F=\tfrac95C+32\).
• Fahrenheit and Kelvin agree at about 574.6 (that is 301.4 °C).
• The Fahrenheit reading is twice the Celsius reading at 160 °C (320 °F).
• Temperature differences: \(\Delta F=1.8\,\Delta C\) and \(\Delta K=\Delta C\). A 1 °C rise is a 1.8 °F rise, not a 33.8 °F rise.

Hold a gas at fixed volume and plot its pressure against Celsius temperature. Every dilute gas gives a straight line. Extend the lines backwards: they all reach zero pressure at the same point, −273.15 °C. This common point is absolute zero, and it does not depend on which gas you use. That is why the gas scale is a good standard.

The Kelvin scale assigns 273.16 K to the triple point of water, where ice, water and vapour coexist. A constant-volume gas thermometer then reads \(T=273.16\ \text{K}\times P/P_{\rm tr}\).

The zeroth law says: if A and B are each in thermal equilibrium with C, then A and B are in thermal equilibrium with each other. It lets us define temperature as the property that is equal for bodies in equilibrium. The thermometer is the body C.

Atoms in a solid vibrate about their equilibrium positions. The potential energy between two neighbouring atoms is not symmetric: it rises steeply if they are pushed together and gently if they are pulled apart. When you heat the solid, the vibrations grow. Because the energy curve is lopsided, the atoms spend more time at larger separations, so the average spacing grows. A perfectly symmetric (spring-like) potential would give no expansion at all.

Take a square plate of side \(L_0\) and heat it by \(\Delta T\).

1. Each side becomes \(L_0(1+\alpha\Delta T)\).
2. Area \(A=L_0^2(1+\alpha\Delta T)^2=A_0\left(1+2\alpha\Delta T+\alpha^2\Delta T^2\right)\).
3. \(\alpha\Delta T\) is about \(10^{-3}\) or smaller, so \((\alpha\Delta T)^2\) is about \(10^{-6}\). Drop it. Then \(\Delta A=2\alpha A_0\Delta T\), so \(\beta=2\alpha\).
4. For a cube, \(V=L_0^3(1+\alpha\Delta T)^3\approx V_0(1+3\alpha\Delta T)\), so \(\gamma=3\alpha\).

Hence \(\alpha:\beta:\gamma=1:2:3\) for an isotropic solid. For a crystal with different coefficients along three axes, the same algebra gives \(\gamma=\alpha_x+\alpha_y+\alpha_z\). The percentage form is quick: if every length grows by p %, area grows by 2p % and volume by 3p %, while density falls by about 3p %.

• Holes and cavities expand exactly like the material that could fill them. A tight ring is heated so it slips over a shaft.
• Density falls: \(\rho=\rho_0/(1+\gamma\Delta T)\approx\rho_0(1-\gamma\Delta T)\).
• Moment of inertia depends on (size)², so \(\Delta I/I=2\alpha\Delta T\). A freely spinning disc that is heated slows down, because \(L=I\omega\) stays constant: \(\Delta\omega/\omega=-2\alpha\Delta T\).
• Bimetallic strip: two metals bonded together bend on heating, with the metal of larger α on the outer (convex) side. Thermostats, fire alarms and flashers use this.

1. Imagine one wall removed. The rod would expand freely by \(\Delta L=L\alpha\Delta\theta\).
2. The wall must push it back by \(\Delta L\) to keep the length fixed.
3. Compressive strain \(=\Delta L/L=\alpha\Delta\theta\), independent of L.
4. Stress \(=Y\times\text{strain}=Y\alpha\Delta\theta\). Force on each wall \(F=YA\alpha\Delta\theta\).

Cooling a rod that is fixed at both ends gives tension of the same size. The elastic energy stored per unit volume is \(\tfrac12\,\text{stress}\times\text{strain}=\tfrac12Y(\alpha\Delta\theta)^2\). This links this chapter with the elasticity chapter.

A pendulum’s period is \(T=2\pi\sqrt{L/g}\). Heating stretches the rod to \(L(1+\alpha\Delta\theta)\), so \(T\) grows to about \(T(1+\tfrac12\alpha\Delta\theta)\).

• Hot: longer period, fewer swings per day, so the clock runs slow and loses time.
• Cold: shorter period, so the clock gains time.
• Time lost or gained per day \(=\tfrac12\alpha\,|\Delta\theta|\times86400\) s.
• If a clock loses at one temperature and gains at another, you can find both the temperature at which it keeps correct time and α.
• A compensated pendulum uses two metals so their expansions cancel: \(L_1\alpha_1=L_2\alpha_2\) keeps the difference \(L_1-L_2\) fixed.

A metal scale is correct at its calibration temperature \(\theta_0\). At a higher temperature each marked centimetre is really a little longer, so the scale reads too small. The true length is \(L_{\rm true}=L_{\rm read}\,[1+\alpha_s(\theta-\theta_0)]\). If the scale and the object are the same material and at the same temperature, the reading stays correct.

Other examples: gaps in railway tracks and bridges, telephone wires that sag in summer, and thick glass that cracks when hot water is poured in, because the inner surface expands before the outer one. Borosilicate (Pyrex) glass has a small α, so it resists such cracking.

Fill a flask of capacity \(V_0\) to the brim with a liquid, then heat both by \(\Delta T\).

1. The liquid’s volume becomes \(V_0(1+\gamma_l\Delta T)\).
2. The flask’s capacity becomes \(V_0(1+\gamma_v\Delta T)\), because the cavity expands like the glass.
3. The overflow is the difference: \(\Delta V=V_0(\gamma_l-\gamma_v)\Delta T\). So the observed (apparent) coefficient is \(\gamma_{\rm app}=\gamma_l-\gamma_v\).

If \(\gamma_l>\gamma_v\) the level rises; if they are equal it stays put; if the vessel expands more, the level falls. When a flask is first dipped into hot water, the level dips for a moment: the glass warms and expands before heat reaches the liquid. Using two vessels of known coefficients, \(\gamma_l=\gamma_{\rm app,1}+\gamma_{v1}=\gamma_{\rm app,2}+\gamma_{v2}\) gives the true coefficient.

Mass is fixed, so \(\rho=\rho_0/(1+\gamma\Delta T)\). For a solid fully immersed in a liquid, upthrust is \(V_s\rho_lg\). On heating, \(V_s\) grows by \(\gamma_s\Delta T\) while \(\rho_l\) falls by \(\gamma_l\Delta T\):

\[ F_B\approx F_{B0}\,[1+(\gamma_s-\gamma_l)\Delta T]. \]

Liquids usually expand much more than solids, so upthrust decreases and the apparent weight of a sunken body increases on heating. A floating body sinks a little deeper in a warmer liquid, because the submerged volume \(m/\rho_l\) must grow.

• From 0 °C to 4 °C, water contracts on warming. Its density is greatest, about 1000 kg/m³, near 4 °C. Above 4 °C it expands normally.
• Ice (about 917 kg/m³) is less dense than water because hydrogen bonds hold the molecules in an open lattice. So ice floats.
• In winter, surface water cools and sinks until the whole lake is near 4 °C. Further cooling makes the surface water lighter, so it stays on top and freezes there.
• Ice is a poor conductor, so it insulates the water below. The bottom stays near 4 °C and aquatic life survives.
• Water pipes burst in severe winters because water expands about 9 % on freezing. Water is also unsuitable as a thermometric liquid near 4 °C, since one volume corresponds to two temperatures.

• Heat capacity \(C=Q/\Delta T\) belongs to a whole object (J/K).
• Specific heat \(c=Q/(m\Delta T)\) belongs to a material (J/kg K).
• Molar heat capacity \(C_m=Mc\) is per mole (J/mol K). For gases it depends on the process, which the thermodynamics chapter uses.
• Water equivalent \(W=mc/c_w\) of a calorimeter is the mass of water that would need the same heat for the same temperature rise. Add it to the water mass in the heat balance.
• Units: 1 cal = 4.186 J (often 4.2 J). Water: \(c_w=4186\ \mathrm{J/kg\,K}=1\ \mathrm{cal/g\,^\circ C}\). Ice: about 2100 J/kg K = 0.5 cal/g °C.

Water has an unusually large specific heat. It warms and cools slowly, which keeps coastal climates mild and makes it a good coolant in car radiators and hot-water bottles.

With no phase change and no heat loss, heat lost by the hot bodies equals heat gained by the cold ones. Solving gives

\[ T_f=\frac{\sum m_ic_iT_i}{\sum m_ic_i}. \]

So the final temperature is an average weighted by heat capacity. It always lies between the extreme starting temperatures, and closer to the body with the larger \(mc\). A simple average is correct only when the heat capacities are equal. Treat the calorimeter as one more body with its own \(mc\).

If \(c\) depends on \(T\), add up small steps: \(Q=\int_{T_1}^{T_2}m\,c(T)\,dT\). For \(c=a+bT\), this gives \(Q=m\left[a(T_2-T_1)+\tfrac b2\left(T_2^2-T_1^2\right)\right]\). The mean specific heat over the range is \(Q/(m\Delta T)\).

Heat a block of ice at −20 °C with a heater of constant power P and plot temperature against time (or heat supplied). The graph has five parts:

1. Ice warms from −20 °C to 0 °C. Slope \(dT/dt=P/(mc_{\rm ice})\).
2. Melting plateau at 0 °C. Ice and water coexist. Length \(=mL_f/P\).
3. Water warms from 0 °C to 100 °C. Slope \(P/(mc_w)\), about half the ice slope because \(c_w\approx2c_{\rm ice}\).
4. Boiling plateau at 100 °C. It is about 6.7 times longer than the melting plateau, because \(L_v/L_f=2.26\times10^6/3.36\times10^5\approx6.7\).
5. Steam warms above 100 °C, again steeply because \(c_{\rm steam}\approx2000\ \mathrm{J/kg\,K}\).

Two rules read any heating curve: a steeper sloping part means a smaller heat capacity, and a longer flat part means a larger latent heat.

During melting or boiling, the supplied energy pulls molecules apart against their attraction. Potential energy rises while average kinetic energy, and so temperature, stays the same. Freezing and condensation release the same latent heat back.

This is why steam burns are worse than boiling-water burns. One gram of steam at 100 °C gives 540 cal as it condenses and up to 100 cal more as the water cools to 0 °C: 640 cal in total, against 100 cal from water at 100 °C.

• Boiling point rises with pressure. A pressure cooker at about 2 atm boils water near 120 °C, so food cooks faster. At high altitude water boils below 100 °C and food cooks slowly.
• The melting point of ice falls when pressure rises, because ice shrinks on melting. A loaded wire passes slowly through an ice block, and the ice refreezes above it (regelation).
• For most substances, such as wax, the solid is denser than the liquid, so their melting point rises with pressure.
• Some solids, like camphor and dry ice, change straight to vapour at ordinary pressure (sublimation). At the triple point of water (273.16 K, about 611 Pa) all three phases coexist.

1. List every body: mass, phase and starting temperature. Include the calorimeter if given.
2. Heat available: how much heat the hot part releases on reaching the transition temperature (0 °C for ice and water, 100 °C for steam and water), including condensation for steam.
3. Heat needed: how much the cold part needs to reach that temperature and change phase completely.
4. Compare and finish: decide the final state, then write one balance equation with the right unknown.

One gram of steam at 100 °C gives 540 cal on condensing and up to 100 cal more on cooling to 0 °C: 640 cal in total. One gram of ice at 0 °C needs 80 cal to melt. So 1 g of steam can melt 8 g of ice, leaving 9 g of water at 0 °C.

For steam + cold water, check the other end. The water can absorb at most \(m_wc_w(100-T_w)\) before it reaches 100 °C. If the steam’s latent heat is larger than that, only part of the steam condenses and the final temperature is 100 °C.

Yes, if the ice is very cold and there is little water. Example: 100 g of ice at −50 °C with 10 g of water at 0 °C (\(c_{\rm ice}=0.5\) cal/g °C).

• Warming all the ice to 0 °C needs \(100\times0.5\times50=2500\) cal.
• Freezing all the water releases only \(10\times80=800\) cal. So all the water freezes.
• Now 110 g of ice settles at T: \(800+10\times0.5\times(0-T)=100\times0.5\times(T+50)\), giving \(T\approx-30.9\ ^\circ\mathrm{C}\).

Fourier’s law in differential form is \(H=\frac{dQ}{dt}=-\kappa A\frac{dT}{dx}\). The minus sign says heat flows down the temperature gradient. When one end of a rod is first heated, each part absorbs heat to warm up. Later the temperatures stop changing: this is the steady state. Then no part stores heat, so the same heat current H crosses every cross-section. For a uniform rod the temperature then falls linearly along its length.

Conduction behaves like current in a resistor:

κ has unit W m⁻¹ K⁻¹ and dimensions [M L T⁻³ K⁻¹].

Series (layers one after another): the same H flows through each layer and the temperature drops add. So \(R=R_1+R_2\). The larger drop occurs across the layer with the larger resistance, which for equal L and A is the poorer conductor. The equivalent conductivity is

\[ \kappa_{\rm eq}=\frac{L_1+L_2}{L_1/\kappa_1+L_2/\kappa_2},\qquad \kappa_{\rm eq}=\frac{2\kappa_1\kappa_2}{\kappa_1+\kappa_2}\ \text{for equal thickness.} \]

Parallel (side by side between the same two faces): each path has the same ΔT and the heat currents add. So \(1/R=1/R_1+1/R_2\) and, for equal lengths, \(\kappa_{\rm eq}=\frac{\kappa_1A_1+\kappa_2A_2}{A_1+A_2}\).

Two identical rods carry 4 times as much heat side by side as end to end, because the resistance changes from 2R to R/2.

Air above a lake is at −θ °C. A layer of ice of thickness x has formed; the water below it is at 0 °C.

1. Heat conducted up through the ice in time dt: \(\frac{\kappa A\theta}{x}\,dt\).
2. This heat comes from water freezing at the bottom of the ice. A new layer dx has mass \(\rho A\,dx\) and releases \(\rho A L\,dx\).
3. Equate: \(\frac{\kappa A\theta}{x}dt=\rho AL\,dx\), so \(dt=\frac{\rho L}{\kappa\theta}x\,dx\).
4. Integrate: \(t=\frac{\rho L}{2\kappa\theta}\left(x_2^2-x_1^2\right)\).

Time grows as the square of thickness. Going from 0 to 2x takes four times as long as 0 to x, so 2x after x takes three times as long. Thick ice insulates the water and slows further freezing.

Ingenhousz’s experiment: identical rods coated with wax are dipped into hot water. In the steady state, the length along which wax melts satisfies \(\ell^2\propto\kappa\), so \(\kappa_1/\kappa_2=\ell_1^2/\ell_2^2\).

Heat a pan of water from below. The bottom layer warms, expands and becomes less dense. Buoyancy lifts it, and cooler, denser water sinks to take its place. The result is a circulating current that carries energy with the moving fluid. This is natural convection.

Natural convection needs gravity, because buoyancy needs weight. In an orbiting spacecraft or a freely falling lift there is no natural convection: a candle flame there becomes a dim sphere, starved of fresh air. Forced convection uses a fan or pump instead: a room fan, a car’s radiator coolant, blood pumped by the heart.

The rate of convective loss from a surface is roughly \(H=hA(T-T_0)\), where h depends on the fluid and how fast it moves. This linear form is one basis of Newton’s law of cooling.

• Sea breeze (day): land heats faster than the sea because of its lower specific heat. Air over the land rises and cooler air from the sea flows in along the surface.
• Land breeze (night): land cools faster, so the sea is warmer. The flow reverses and air moves from land to sea.
• Trade winds and monsoons: large-scale convection driven by unequal heating of the Earth.
• Heating elements sit at the bottom of a kettle. Freezers and air conditioners sit high up, so cooled air sinks through the space.
• A chimney works because hot gases inside are lighter than outside air.

Radiation falling on a surface is partly absorbed (fraction a), reflected (r) and transmitted (t), with \(a+r+t=1\). A black body has \(a=1\). Emissivity \(\varepsilon\) compares a surface’s emission with a black body’s at the same temperature.

Kirchhoff’s law: at a given temperature and wavelength, the ratio of emissive power to absorptivity is the same for all bodies and equals the black-body emissive power. So \(\varepsilon=a\): good absorbers are good emitters, and good reflectors are poor emitters.

• A plate with a black pattern on white china, heated in a dark room, shows the black pattern glowing brighter.
• A thermos flask has silvered walls: they reflect radiation and emit little.
• Light-coloured clothes are cooler in summer because they absorb less sunlight.
• A small hole in a blackened cavity behaves as a near-perfect black body (Ferry’s black body). Inside a furnace, all objects glow alike and their details vanish.

A body emits \(\varepsilon\sigma AT^4\) and absorbs \(\varepsilon\sigma AT_0^4\) from surroundings at \(T_0\). Even at \(T=T_0\) it still emits; it simply absorbs the same amount, so the net power is zero (Prevost’s theory of exchanges).

The rate of fall of temperature is \(-\frac{dT}{dt}=\frac{\varepsilon\sigma A}{mc}(T^4-T_0^4)\). Two common comparisons:

• Solid spheres of the same material at the same T: \(A/m=\frac{4\pi r^2}{\rho\cdot\frac43\pi r^3}=\frac{3}{\rho r}\), so the cooling rate \(\propto1/r\). The smaller sphere cools faster.
• A hollow and a solid sphere of the same radius and material: same emission at first, but the hollow one has less mass, so it cools faster.

A star of radius R radiates a total power (luminosity) \(L=4\pi R^2\sigma T^4\). For a small excess, \(T^4-T_0^4\approx4T_0^3(T-T_0)\), which leads to Newton’s law of cooling in the next section.

Plot the power emitted per unit wavelength against wavelength (see the figure). At a higher temperature:

• the curve lies higher at every wavelength, so curves at different temperatures never cross;
• the peak moves to a shorter wavelength, \(\lambda_m\propto1/T\);
• the area under the curve, which is the total power per unit area, grows as \(T^4\).

So heated iron glows dull red, then orange, then yellow-white. Blue stars are hotter than red stars. The Sun peaks near 500 nm, giving about 5800 K. The human body at 310 K peaks near 9.4 μm, in the infrared, which is why thermal cameras work in the infrared.

A body at T in surroundings at \(T_0\) loses heat by radiation at the net rate \(\varepsilon\sigma A(T^4-T_0^4)\).

1. Factorise: \(T^4-T_0^4=(T^2+T_0^2)(T+T_0)(T-T_0)\).
2. If T is close to \(T_0\), the first two brackets are about \(2T_0^2\times2T_0=4T_0^3\).
3. So the loss rate is about \(4\varepsilon\sigma AT_0^3\,(T-T_0)\): proportional to the excess temperature.
4. Divide by the heat capacity mc: \(\frac{dT}{dt}=-k(T-T_0)\) with \(k=\frac{4\varepsilon\sigma AT_0^3}{mc}\).

So k is larger for a larger surface and a smaller heat capacity. The law holds only for small excess temperatures (a few tens of kelvin), though forced convection also gives a roughly linear loss.

Exact form. Integrating gives \(\ln\frac{T_1-T_0}{T_2-T_0}=kt\). Equal time intervals give equal fractional drops in the excess, as in the figure.

Average form (the one used in most NEET problems). Replace the changing rate by its value at the mean temperature of the interval:

\[ \frac{T_1-T_2}{t}=K\left[\frac{T_1+T_2}{2}-T_0\right]. \]

The two forms agree closely when the drop \(T_1-T_2\) is small compared with the excess. Use the average form unless the question asks for the logarithmic one or gives ln values.

• T against t: an exponential decay that approaches \(T_0\) and never crosses it. Steepest at the start.
• \(\ln(T-T_0)\) against t: a straight line of slope −k.
• Rate of cooling against \((T-T_0)\): a straight line through the origin.
• Rate of cooling against T: a straight line that meets the T axis at \(T_0\).

Specific heat by cooling. Put equal volumes of water and another liquid in identical calorimeters and let them cool through the same range. Same surface and same temperatures give the same rate of heat loss, so \(\frac{m_1c_1+W}{t_1}=\frac{m_2c_2+W}{t_2}\), where W is the calorimeter’s heat capacity.