A steel rod is clamped rigidly between two walls and heated by 50 K. With Y = 2 × 10¹¹ Pa and α = 1.2 × 10⁻⁵ K⁻¹, the thermal stress in the rod is:

It depends on the length of the rod

Thermal Properties of Matter — Expansion, Calorimetry, Conduction and Radiation

The thermodynamics chapter follows energy through a gas; this one follows heat through ordinary matter. Four ideas carry it: bodies grow when warmed, and by a predictable fraction; heat lost by one body in a mixture is heat gained by another, with phase changes taking their share at constant temperature; heat flows through a slab in proportion to the temperature difference, exactly as current flows through a resistor; and every surface radiates, at a rate that climbs as the fourth power of its absolute temperature. Each idea is one equation, and the harder problems come from combining two of them — a rod that conducts and radiates, a mixture where the ice may not all melt.

Thermal Expansion — a Fixed Fraction per Kelvin

Warm a rod and its atoms vibrate about positions slightly farther apart, because the interatomic potential is steeper on the compression side than on the stretching side. The result, for small temperature changes, is a length change proportional to both the length and the temperature rise: ΔL = αL ΔT, where α is the coefficient of linear expansion. An area grows as ΔA = βA ΔT and a volume as ΔV = γV ΔT. For an isotropic solid every direction expands alike, so to first order β = 2α and γ = 3α — a cube of side L becomes L³(1 + αΔT)³ ≈ L³(1 + 3αΔT).

Two consequences are asked again and again. A hole expands as if it were made of the material around it — a heated plate with a circular hole has a larger hole, not a smaller one, because every length in the plate, including the hole's diameter, scales by the same factor. And a rod prevented from expanding develops a compressive thermal stress: the strain it is denied is αΔT, so the stress is Y α ΔT, independent of the rod's length or cross-section.

⚠️ Water between 0 °C and 4 °C

Water contracts when warmed from 0 °C to 4 °C and is densest at about 4 °C. That is why a lake freezes from the top: the 4 °C water sinks and the colder, lighter water stays at the surface to freeze. Any problem that assumes "warmer means bigger" for water near freezing is set to catch you.

Calorimetry and Latent Heat — Heat Lost Equals Heat Gained

Raising a mass m by ΔT without a change of phase takes Q = m c ΔT, with c the specific heat capacity. A change of phase takes Q = m L at constant temperature, with L the latent heat — 80 cal g⁻¹ (about 336 J g⁻¹) to melt ice and about 540 cal g⁻¹ to boil water. In an insulated mixture the principle of calorimetry says that heat given out by the hot parts equals heat taken in by the cold parts.

🎯 Worked reasoning: check the phase change before you write one equation

Mix 50 g of ice at 0 °C with 100 g of water at 30 °C. Writing one equation with a single final temperature T gives 50 × 80 + 50 × T = 100 × (30 − T), so T = −6.7 °C — impossible, because ice-water cannot be below 0 °C. The honest method is to compare budgets first: the water can give at most 100 × 30 = 3000 cal by cooling to 0 °C, while melting all the ice needs 4000 cal. So the mixture ends at 0 °C with 3000/80 = 37.5 g melted and 12.5 g of ice left. A negative or impossible answer is the signal to switch methods, not to report it.

Conduction — Heat Flow Behaves Like Current

Through a slab of thickness L, area A and thermal conductivity k, with faces held at T₁ and T₂, the steady rate of heat flow is H = dQ/dt = kA(T₁ − T₂)/L (Fourier's law). Read it as Ohm's law: the temperature difference is the "voltage", H is the "current", and R_th = L/(kA) is the thermal resistance. Slabs one after another add like resistors in series (same H through each); slabs side by side add like resistors in parallel (same ΔT across each).

The thermal–electrical analogy
Heat conduction	Electric circuit
Temperature difference ΔT	Potential difference ΔV
Heat current H	Electric current I
R_th = L/(kA)	R = ρL/A = L/(σA)
Junction temperature between two slabs	Potential at the node between two resistors

🎯 Worked reasoning: the junction temperature

Two rods of equal length and area, with conductivities 2k and k, are joined end to end; the free end of the better conductor is at 100 °C and the other at 0 °C. In series the same H flows through both, so 2k(100 − T) = k(T − 0), giving T = 200/3 ≈ 66.7 °C. The junction sits nearer the hot end, because the better conductor needs a smaller temperature drop to carry the same heat — exactly as a smaller resistor takes a smaller share of the voltage.

Convection, Radiation and Newton's Law of Cooling

Convection carries heat by bulk motion of a fluid — warm fluid is less dense and rises — so it needs gravity and a fluid, and it is treated only qualitatively. Radiation needs no medium at all. A body at absolute temperature T with surface area A and emissivity e radiates power P = eσAT⁴ (Stefan–Boltzmann law, σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴), and while surrounded by an environment at T₀ it absorbs eσAT₀⁴, so the net loss is eσA(T⁴ − T₀⁴). A perfect black body has e = 1.

Kirchhoff's law: at a given temperature and wavelength, the ratio of emissive power to absorptive power is the same for every body and equals the black body's emissive power — so a good absorber is a good emitter. Wien's displacement law: the wavelength at which a black body's spectrum peaks shifts inversely with temperature, λ_max T = b ≈ 2.9 × 10⁻³ m K; a surface at about 5800 K peaks near 500 nm. Newton's law of cooling: for a small excess temperature, the rate of cooling is proportional to the excess over the surroundings, dT/dt = −k(T − T₀), so the excess decays exponentially, (T − T₀) = (T₁ − T₀)e^(−kt).

⚠️ Kelvin, always, inside T⁴

Doubling a temperature from 27 °C to 54 °C does not multiply the radiated power by 16: in kelvin it goes from 300 K to 327 K, a factor of (327/300)⁴ ≈ 1.41. Only absolute temperature belongs in Stefan's and Wien's laws. Celsius is safe only where a temperature difference appears, as in conduction and Newton's law.

Key takeaways

ΔL = αLΔT, and for an isotropic solid β = 2α and γ = 3α; a hole in a heated plate grows, and a clamped rod feels thermal stress YαΔT.
In calorimetry, compare heat budgets before writing a single-temperature equation; if the ice cannot all melt, the mixture stays at 0 °C.
Conduction is Ohm's law in disguise: H = ΔT/R_th with R_th = L/(kA), and series and parallel slabs combine like resistors.
Radiated power is eσAT⁴ with T in kelvin; good absorbers are good emitters; λ_max T ≈ 2.9 × 10⁻³ m K.
Newton's law of cooling makes the excess temperature decay exponentially; equal times cut the excess by equal factors.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

An isotropic solid has coefficient of linear expansion α. To first order, its coefficient of volume expansion is:
1. 3α
2. α
3. 2α
4. α³
Show answer
Answer: A — 3α
Each of the three dimensions grows by the factor (1 + αΔT), so the volume grows by (1 + αΔT)³ ≈ 1 + 3αΔT, giving γ = 3α. 2α is the area coefficient β, and α³ cubes the coefficient instead of the growth factor — a small number made much smaller.
A steel rod is clamped rigidly between two walls and heated by 50 K. With Y = 2 × 10¹¹ Pa and α = 1.2 × 10⁻⁵ K⁻¹, the thermal stress in the rod is:
1. 2.4 × 10⁶ Pa
2. 1.2 × 10⁸ Pa
3. 6 × 10⁻⁴ Pa
4. It depends on the length of the rod
Show answer
Answer: B — 1.2 × 10⁸ Pa
The denied strain is αΔT = 1.2 × 10⁻⁵ × 50 = 6 × 10⁻⁴, so the stress is Y αΔT = 2 × 10¹¹ × 6 × 10⁻⁴ = 1.2 × 10⁸ Pa. The length cancels — a longer rod would expand more, but it is the strain, a ratio, that sets the stress. 6 × 10⁻⁴ is the strain itself, not the stress.
100 g of water at 60 °C is mixed with 50 g of water at 30 °C in an insulated vessel of negligible heat capacity. The final temperature is:
1. 45 °C
2. 40 °C
3. 50 °C
4. 55 °C
Show answer
Answer: C — 50 °C
Heat lost = heat gained: 100(60 − T) = 50(T − 30), so 6000 + 1500 = 150T and T = 50 °C — the mass-weighted mean. 45 °C is the plain average of the two temperatures, which would be right only for equal masses.
With c(ice) = 0.5 cal g⁻¹ °C⁻¹, L(fusion) = 80 cal g⁻¹ and c(water) = 1 cal g⁻¹ °C⁻¹, the heat needed to turn 10 g of ice at −10 °C into water at 20 °C is:
1. 1000 cal
2. 300 cal
3. 1100 cal
4. 1050 cal
Show answer
Answer: D — 1050 cal
Three stages: warm the ice to 0 °C, 10 × 0.5 × 10 = 50 cal; melt it, 10 × 80 = 800 cal; warm the water to 20 °C, 10 × 1 × 20 = 200 cal. Total 1050 cal. Using the water's specific heat for the ice stage gives 1100 cal, and leaving out the latent heat gives a figure that ignores the largest term of all.
Rods A and B have equal length and cross-section; A has twice the thermal conductivity of B. They are joined end to end, with the free end of A held at 100 °C and that of B at 0 °C. In the steady state the junction temperature is:
1. 50 °C
2. 200/3 °C
3. 100/3 °C
4. 75 °C
Show answer
Answer: B — 200/3 °C
In series the heat current is the same in both rods: 2k(100 − T) = k(T − 0), so 200 = 3T and T = 200/3 ≈ 66.7 °C. The better conductor has the smaller thermal resistance and so takes the smaller share of the 100 °C drop. 100/3 °C puts the junction near the cold end, which is what you get by attaching the conductivities to the wrong rods.
A black body radiates power P at absolute temperature T. If its absolute temperature is doubled, with surroundings negligible, it radiates:
1. 2P
2. 4P
3. 16P
4. 8P
Show answer
Answer: C — 16P
Stefan–Boltzmann gives P ∝ T⁴, so doubling T multiplies P by 2⁴ = 16. 4P would be a T² law and 2P a linear one. The doubling here is of absolute temperature; doubling a Celsius reading would not give 16.
Taking Wien's constant as 2.9 × 10⁻³ m K, a black-body surface at 5800 K has its spectral peak at a wavelength of about:
1. 500 nm
2. 5 μm
3. 50 nm
4. 17 m
Show answer
Answer: A — 500 nm
λ_max = b/T = 2.9 × 10⁻³/5800 = 5 × 10⁻⁷ m = 500 nm, in the visible. The factor-of-ten options come from slipping a power of ten in the division, and 17 m multiplies b by T instead of dividing (2.9 × 10⁻³ × 5800 ≈ 16.8).
A body cools from 80 °C to 60 °C in 10 minutes in surroundings at 20 °C. Using the exact exponential form of Newton's law of cooling, the time it then takes to cool from 60 °C to 40 °C is about:
1. 10 min
2. 15 min
3. 20 min
4. 17.1 min
Show answer
Answer: D — 17.1 min
The excess over 20 °C falls from 60 to 40 in 10 min, a factor of 2/3, so e^(−10k) = 2/3 and k = ln(1.5)/10. Cooling from 60 °C to 40 °C takes the excess from 40 to 20, a factor of 1/2, so t = ln 2/k = 10 × 0.693/0.405 ≈ 17.1 min. Answering 10 min assumes equal temperature drops take equal times; the body cools more slowly as it nears the surroundings.