Thermal Properties of Matter: Full Notes

10.1 Temperature and Thermal Expansion

Temperature is simply a measure of the degree of hotness of a body, nothing more mysterious than that. What makes it useful is the zeroth law of thermodynamics: two bodies, each separately in thermal equilibrium with a third, are in thermal equilibrium with each other. That shared property is what a thermometer actually reads, and it is why a single number on a scale can meaningfully compare the hotness of a cup of tea and a bar of iron.

Three scales matter here. The Celsius scale fixes 0°C at the ice point and 100°C at the steam point. The Fahrenheit scale fixes the same two points at 32°F and 212°F, giving 180 Fahrenheit degrees to every 100 Celsius degrees. The Kelvin scale is the SI, absolute scale, with the same size degree as Celsius but its zero at absolute zero rather than the ice point:

T(\text{K}) = T(^\circ\text{C}) + 273.15, \qquad T(^\circ\text{F}) = \dfrac{9}{5}T(^\circ\text{C}) + 32

10.1.1 Linear, area and volume expansion

Nearly every solid expands on heating, because the average separation between its atoms grows as they vibrate more vigorously about their lattice sites. For a rod of length $L$ , the fractional change in length per unit rise in temperature is its coefficient of linear expansion $\alpha$ :

\Delta L = L\,\alpha\,\Delta T

The same idea applies to an area $A$ (coefficient $\beta$ ) and a volume $V$ (coefficient $\gamma$ ):

\Delta A = A\,\beta\,\Delta T, \qquad \Delta V = V\,\gamma\,\Delta T

For an isotropic solid, one that expands equally in every direction, these three are not independent. A square of side $L$ has area $L^2$ , and expanding each side by the same small fraction stretches the area by roughly twice that fraction; a cube of volume $L^3$ stretches by roughly three times that fraction. Dropping terms of order $\alpha^2$ and smaller (valid since $\alpha$ is typically only $10^{-5}$ per kelvin) gives the standard rule of thumb:

Expansion coefficient ratio

\textcolor{#e08a1e}{\alpha : \beta : \gamma \;\approx\; 1 : 2 : 3}

Worked example

Thermal expansion of a steel rod

A steel surveyor’s rod is exactly 2.000 m long at 20°C. Taking $\alpha_{\text{steel}} = 1.2\times10^{-5}\text{ K}^{-1}$ , find its length at 100°C.

The temperature rise is 80 K, so:

\Delta L = L\,\alpha\,\Delta T = 2.000\times(1.2\times10^{-5})\times80 = 1.92\times10^{-3}\text{ m} = 1.92\text{ mm}

so the rod reads $\textcolor{#e08a1e}{2.00192\text{ m}}$ at 100°C, nearly 2 mm longer than at 20°C, which is exactly why precision instruments and railway tracks are calibrated for, or built to tolerate, a stated temperature range.

10.2 Specific Heat Capacity and Calorimetry

Raising a body’s temperature takes heat, and how much depends on the substance and the mass. The specific heat capacity $c$ of a substance is the heat required to raise the temperature of unit mass by one kelvin, so for a mass $m$ absorbing heat $\Delta Q$ with a temperature change $\Delta T$ :

\Delta Q = m\,c\,\Delta T

Chemists more often quote the molar specific heat $C = Mc$ , heat per mole rather than per kilogram, where $M$ is the molar mass; both versions describe the exact same physical heat capacity, just per different amounts of substance. Water’s specific heat, 4186 J/(kg·K), is unusually large among common substances, which is precisely why large bodies of water resist temperature swings and moderate coastal climates.

Calorimetry is the practical use of this idea: when a hot body and a cold body are placed in contact inside an insulated calorimeter and left alone, no heat escapes to the surroundings, it only moves from the hotter object to the colder one until both reach a common final temperature. This is the principle of calorimetry:

Principle of calorimetry

\textcolor{#e08a1e}{\text{Heat lost by the hot body}} = \textcolor{#e08a1e}{\text{Heat gained by the cold body}}

Worked example

Finding an unknown specific heat by mixing

A 200 g piece of metal at 150°C is dropped into 500 g of water at 25°C inside a calorimeter whose own heat capacity can be ignored. The mixture settles at a final temperature of 30°C. Find the specific heat capacity of the metal. ( $c_{\text{water}} = 4186\text{ J/(kg·K)}$ )

The metal cools by 150 − 30 = 120 K while the water warms by 30 − 25 = 5 K. Setting heat lost equal to heat gained:

m_{\text{metal}}\,c_{\text{metal}}\,(150-30) = m_{\text{water}}\,c_{\text{water}}\,(30-25)

0.2\times c_{\text{metal}}\times120 = 0.5\times4186\times5 = 10465\text{ J}

c_{\text{metal}} = \dfrac{10465}{24} \approx \textcolor{#e08a1e}{436\text{ J/(kg·K)}}

close to the known value for iron (about 450 J/(kg·K)), which is exactly how this kind of experiment is used in practice, to identify or verify an unknown material from a simple mixing measurement.

10.3 Change of State and Latent Heat

Heating a solid does not always raise its temperature. At a melting or boiling point, the incoming energy is spent entirely on pulling molecules apart against the intermolecular forces holding the solid or liquid together, rearranging the bonds rather than speeding the molecules up, so the temperature stays fixed until the change of state is complete. The heat needed per unit mass to carry out this change, with no temperature change at all, is the latent heat $L$ :

Q = mL

For water at 1 atmosphere, the latent heat of fusion (ice to water, at 0°C) is $L_f = 333\text{ kJ/kg}$ , and the latent heat of vaporisation (water to steam, at 100°C) is $L_v = 2260\text{ kJ/kg}$ .

Plotting temperature against total heat added to a fixed mass of ice, starting well below 0°C and finishing above 100°C, produces the chapter’s signature heating curve: the temperature climbs steadily while the sample is purely solid, then flatlines at 0°C while it melts, climbs again through the liquid range, flatlines again at 100°C while it boils, and finally climbs once more through the vapour range. Both flat stretches are regions where heat keeps flowing in but the thermometer refuses to move, because every joule is going into breaking bonds, not into the kinetic energy that temperature measures.

Worked example

Comparing the two plateaus

For 1 kg of water, how much heat does each plateau of the heating curve represent, and which is wider?

Q_{\text{melt}} = mL_f = 1\times333 = 333\text{ kJ}

Q_{\text{boil}} = mL_v = 1\times2260 = 2260\text{ kJ}

\dfrac{Q_{\text{boil}}}{Q_{\text{melt}}} = \dfrac{2260}{333} \approx \textcolor{#e08a1e}{6.8}

Turning water into steam takes nearly 6.8 times more heat than melting the same mass of ice, which is exactly why the boiling plateau is drawn so much wider than the melting plateau on the heating curve, and why steam burns are so much more dangerous than hot-water burns: every gram of steam that condenses on skin dumps roughly seven times the energy that the same gram of ice melting would absorb.

10.4 Heat Transfer: Conduction

Heat moves from a hotter region to a colder one by three distinct mechanisms, conduction, convection and radiation. Conduction is heat transfer through a material without any bulk motion of the material itself: faster-vibrating atoms at the hot end jostle their neighbours, passing energy along, and in metals free electrons carry it even faster. Consider a slab of cross-sectional area $A$ and thickness $x$ , with its two faces held at steady temperatures $T_1 > T_2$ . In this steady state, the rate of heat flow through the slab is:

Steady-state conduction

\textcolor{#e08a1e}{\dfrac{dQ}{dt}} = \dfrac{kA(T_1-T_2)}{x}

where $k$ , the thermal conductivity, is a property of the material alone (SI unit W/(m·K)), large for metals like copper and silver and tiny for insulators like wood, air or thermocole. A good conductor needs only a small temperature difference to pass a given amount of heat; a poor one needs a large difference for the same flow.

10.4.1 Compound slabs and thermal resistance

It helps to define a slab’s thermal resistance $R = x/(kA)$ , so the conduction law reads $dQ/dt = \Delta T/R$ , exactly like Ohm’s law with temperature difference in place of voltage and heat current in place of electric current. For two slabs of different materials placed one after the other (same area $A$ , the same steady heat current passing through both in series), the resistances simply add:

R_{\text{eq}} = R_1 + R_2 = \dfrac{x_1}{k_1A} + \dfrac{x_2}{k_2A}

so a thin layer of a very poor conductor, like the insulation in a wall or the air gap in a double-glazed window, can dominate the total resistance and choke the heat flow even when every other layer conducts well.

NCERT Exercise 10.9

A brass boiler has a base area of 0.15 m² and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler base. Thermal conductivity of brass = 109 J s⁻¹ m⁻¹ K⁻¹; heat of vaporisation of water = 2256 × 10³ J/kg.

Solution

The flame must supply, by conduction through the base, exactly the heat needed to vaporise water at that rate. The mass flow rate is 6.0 kg/min = 0.1 kg/s, so the required heat current is:

\dfrac{dQ}{dt} = \left(\dfrac{dm}{dt}\right)L_v = 0.1\times(2256\times10^3) = 2.256\times10^5\text{ W}

Using the steady-state conduction law with this as the heat current through the brass base, and solving for the temperature difference across it:

\Delta T = \dfrac{(dQ/dt)\,x}{kA} = \dfrac{(2.256\times10^5)\times0.01}{109\times0.15} \approx 138^\circ\text{C}

Since the inner, water-facing surface is pinned at the boiling point, 100°C, the flame side of the base must be hotter than that by this amount:

T_{\text{flame side}} = 100 + 138 \approx \textcolor{#e08a1e}{238^\circ\text{C}}

10.5 Convection and Radiation

Convection is heat transfer by the actual bulk movement of a heated fluid. Heated fluid near a source expands, becomes less dense, and rises, while cooler, denser fluid sinks to take its place, setting up a circulating current that carries heat along with the moving material itself. This is natural convection, responsible for land and sea breezes, the rolling currents inside a pot of boiling water, and the atmosphere’s weather patterns; pushing the fluid along mechanically, with a fan or a pump, is forced convection, as in a car’s radiator or the human circulatory system distributing body heat.

Radiation needs no medium at all. Every body, simply by virtue of being above absolute zero, continuously emits electromagnetic radiation, which is how the Sun’s heat crosses empty space to reach the Earth. A hotter body radiates more total power, and far more steeply than linearly: an ideal radiator (a black body) obeys the Stefan–Boltzmann law,

E = \sigma A T^4

with $\sigma = 5.67\times10^{-8}\text{ W/(m}^2\text{K}^4)$ the Stefan–Boltzmann constant, so doubling a body’s absolute temperature multiplies its radiated power by a factor of 16.

10.5.1 Newton's law of cooling

A hot object loses heat to its surroundings mainly by convection and radiation together, and tracking the full $T^4$ dependence of radiation is more detail than most situations need. Newton’s law of cooling is the useful approximation: when a body’s excess temperature over its surroundings is small, its rate of cooling is approximately proportional to that excess,

Newton's law of cooling

\textcolor{#e08a1e}{-\dfrac{dT}{dt}} = k\,(T - T_0)

where $T_0$ is the (constant) surrounding temperature and $k$ depends on the body’s area, surface properties and heat capacity. It is only an approximation, obtained by linearising the true radiative loss around $T\approx T_0$ , and it holds only for a small temperature difference from the surroundings; for a body much hotter than its surroundings, the true $T^4$ behaviour takes over and cooling proceeds noticeably faster than this linear law predicts.

NCERT Exercise 10.10

A pan filled with hot food cools from 94°C to 86°C in 2 minutes when the room temperature is 20°C. How long will it take for the same pan to cool from 71°C to 69°C?

Solution

Newton’s law is applied in its usual finite-difference form, replacing the instantaneous rate with the average temperature over each small interval. For the first cooling, the average temperature is (94+86)/2 = 90°C, an excess of 70°C over the room:

\dfrac{94-86}{120} = k\,(90-20) \;\Rightarrow\; k = \dfrac{8}{120\times70} = \dfrac{1}{1050}\text{ s}^{-1}

For the second interval the average temperature is (71+69)/2 = 70°C, an excess of 50°C, and the same constant $k$ applies:

\dfrac{71-69}{t} = k\,(70-20) = \dfrac{1}{1050}\times50

t = \dfrac{2\times1050}{50} = \textcolor{#e08a1e}{42\text{ s}}

A much shorter time than the first interval, exactly as expected: the pan is now barely above room temperature, so it has far less driving temperature difference left to cool with.