Thermal Expansion and Quantity of Heat

Heat and Thermodynamics carries 7 of the 50 Physics questions (official MEC weightage), and we estimate about 2 of them come from this chapter. The marks here are almost all numerical or definitional: a coefficient of expansion and its relation to the other two, a liquid that overflows a vessel, a mixture that settles at a final temperature, an ice–water–steam sum, or a clock that loses seconds on a hot day. Every one of them is a one-line formula provided the phase bookkeeping is done in the right order. That order is what this chapter teaches.

9.1 Linear, superficial and cubical expansion and their relation

Almost every substance expands when heated at constant pressure. The atoms vibrate about fixed mean positions; the vibration is not symmetric about the mean position — the repulsive side of the potential is steeper than the attractive side — so the average separation grows as the amplitude grows. Nothing in the solid is pushed apart by a "force of heat"; the spacing simply increases.

The three coefficients

All three have SI unit K⁻¹ (numerically the same as °C⁻¹, because a temperature difference is the same on both scales). They are properties of the material, not of the size or shape of the body: a long rod and a short rod of copper have the same α.

Quantity	Symbol	Formula	Unit
Change in length	\Delta L	L\alpha\Delta T	m
Change in area	\Delta A	A\beta\Delta T	m²
Change in volume	\Delta V	V\gamma\Delta T	m³
New length	L'	L(1+\alpha\Delta T)	m

Typical values at room temperature (K⁻¹):

Material	\alpha	Material	\alpha
Quartz (fused)	0.4\times10^{-6}	Copper	1.7\times10^{-5}
Invar	1.2\times10^{-6}	Brass	1.9\times10^{-5}
Glass	8\times10^{-6}	Aluminium	2.4\times10^{-5}
Steel, iron	1.2\times10^{-5}	Lead	2.9\times10^{-5}

Two things to read off this table. Invar was invented for pendulum rods and tape measures precisely because its α is ten times smaller than steel's. And liquids and gases expand far more than solids: for mercury \gamma = 1.8\times10^{-4} K⁻¹, for water about 2.1\times10^{-4} K⁻¹, for air at constant pressure \gamma = 1/273 \approx 3.7\times10^{-3} K⁻¹. A gas is some twenty times more expansive than a liquid and a few hundred times more than a solid.

Why β = 2α and γ = 3α

The relation holds for any linear dimension of the body, including a hole. A plate with a circular hole is, geometrically, the same as a plate with a disc of the same material filling the hole — so on heating the disc grows and the hole grows with it. The hole does not close up. This single fact is asked in some form in almost every paper.

Bimetallic strip

Two different metals — usually brass and iron — are riveted or welded along their whole length. Brass has the larger α (about 1.9\times10^{-5} against 1.2\times10^{-5} K⁻¹).

On heating, the brass side becomes longer than the iron side, so the strip can only stay joined by bending into an arc in which the brass is on the outside (the longer path). On cooling the brass contracts more, so it ends up on the inside. The strip therefore bends towards the metal with the smaller α when heated.

Uses: the thermostat in an electric iron, a geyser or a refrigerator (the strip curls, breaks a contact, the heater switches off, and it straightens again when the temperature falls); the bimetallic thermometer; and temperature compensation in watches.

Thermal stress

If a rod is clamped at both ends, it cannot expand, and the "lost" expansion appears as an elastic strain:

where Y is Young's modulus. The stress is compressive on heating (the rod would have grown, so it is squeezed) and tensile on cooling. This is why concrete roads are laid in slabs with tar-filled gaps, why railway rails are welded with expansion joints, and why a glass tumbler cracks when boiling water is poured into it: the inner surface tries to expand against a cold, unyielding outer surface.

9.2 Real and apparent expansion of liquids

A liquid has no fixed shape, so it has only one coefficient — the cubical one, \gamma. (Writing "the linear expansion of water" is meaningless.) Its expansion is measured by watching the level in a narrow stem, and that is where the trouble starts: the vessel is expanding too, so the scale you read against is itself growing.

Apparent and real expansion

Let the vessel be filled to the brim at the lower temperature, with volume V and coefficient \gamma_v; let the liquid's real coefficient be \gamma_r.

Vessel capacity at the higher temperature: V_v = V(1+\gamma_v\Delta T). Liquid volume at the higher temperature: V_l = V(1+\gamma_r\Delta T).

The liquid that overflows is the difference:

But the overflow is, by definition, the apparent expansion V\gamma_a\Delta T. Comparing the two:

So the real coefficient is always larger than the apparent one, and the gap is exactly the expansion of the vessel. For a glass vessel \gamma_v = 3\alpha_{glass} \approx 2.4\times10^{-5} K⁻¹, which is small beside mercury's 1.8\times10^{-4} but not negligible — it is about 13 % of it.

Liquid in a glass vessel	\gamma_r (K⁻¹)	\gamma_a = \gamma_r - \gamma_v (K⁻¹)
Mercury	1.82\times10^{-4}	1.58\times10^{-4}
Water (around 20 °C)	2.1\times10^{-4}	1.86\times10^{-4}
Alcohol	1.1\times10^{-3}	1.08\times10^{-3}
Paraffin	9.0\times10^{-4}	8.8\times10^{-4}

Why the level first falls, then rises

Warm a flask of coloured water gently and watch the top of the liquid in the stem. It dips slightly, then climbs steadily past its starting mark.

The reason is a race between two expansions that start at the same moment but do not have the same size. The glass in contact with the warming liquid heats up first and quickly, so the flask's capacity grows early. The liquid as a whole is a poor conductor and warms slowly, so for the first few seconds its own expansion is still tiny. Capacity up, liquid not yet up → level down. Once the bulk liquid warms, its much larger \gamma takes over and the level rises.

A second, quite different cause of a dip is anomalous expansion: between 0 °C and 4 °C water contracts as it warms, so its level genuinely falls.

Anomalous expansion of water

Water is not a normal liquid. From 0 °C to 4 °C it contracts; above 4 °C it expands in the ordinary way. Its density therefore peaks at 4 °C:

Consequences you can be asked to explain:

• Ice floats, because ice (917 kg m⁻³) is less dense than water.
• A pond freezes from the top down. The surface water cools to 4 °C and sinks; below 4 °C it is lighter and stays on top, where it freezes. The 4 °C water at the bottom lets fish survive the winter.
• Water in a closed pipe bursts when it freezes: the 8 % expansion is resisted by the pipe, and the pressure rise is enormous.

Density falls as the temperature rises. From V = V_0(1+\gamma\Delta T) and \rho = m/V,

so a rise of \Delta T lowers the density by the fraction \gamma\Delta T. A body's relative density falls on heating, and a solid floats a little deeper in hot water than in cold.

9.3 Heat capacity, specific heat capacity and the method of mixtures

The three quantities

So Q = mc\Delta T = C\Delta T, and C = mc. A 2 kg block of copper has twice the heat capacity of a 1 kg block, but the same specific heat capacity.

Substance	c (J kg⁻¹ K⁻¹)	Substance	c (J kg⁻¹ K⁻¹)
Water	4200	Brass	380
Ice	2100	Copper	390
Steam	2010	Iron, steel	480
Kerosene	2100	Aluminium	900
Mercury	140	Lead	130

Water's specific heat is unusually large — about ten times that of copper. That one number explains why water is used as a coolant in vehicle radiators, why a hot-water bottle stays warm for hours, why coastal climates are milder than inland ones, and why a kettle of water takes so long to boil. It is also why water is the standard calorimetric liquid: a given mass of water changes temperature slowly, so a mixture experiment is easy to time.

Water equivalent

A calorimeter is a metal cup, and it takes heat too. Rather than carry its mass and specific heat through every calculation, we convert it into the mass of water that would take the same heat for the same temperature rise:

A copper calorimeter of mass 0.10 kg and c = 390 J kg⁻¹ K⁻¹ has water equivalent W = 0.10\times390/4200 = 0.0093 kg = 9.3 g. Its heat capacity is 9.3\times10^{-3}\times4200 = 39 J K⁻¹.

The method of mixtures

Put a hot solid of mass m_1, specific heat c_1, at temperature \theta_1 into water of mass m_2 at \theta_2, inside a calorimeter of water equivalent W. If the final temperature is \theta (with \theta_2 < \theta < \theta_1):

Every bracket is positive: a hot body falls from \theta_1 to \theta, so its change is \theta_1-\theta; a cold body rises from \theta_2 to \theta, so its change is \theta-\theta_2. Writing the two brackets the same way round is the source of most sign errors in this topic.

Three ways a mixture answer goes wrong

1. Ignoring the calorimeter. Its water equivalent must be added to the water mass. Leaving it out always makes the hot body's specific heat come out too low (too little heat is credited to the cold side).
2. Sign flip. Use (\theta_1-\theta) for the hot side and (\theta-\theta_2) for the cold side. A negative answer means you reversed one of them.
3. Stopping at 0 °C. If ice is involved, check first whether all the ice can melt with the heat available. If it cannot, the final temperature is 0 °C with some ice left, and the water term must not be used at all.

Newton's law of cooling

Since dQ = -mc\,d\theta for the body, the same law can be written as a statement about the rate of fall of temperature:

Integrating from \theta_0 at t = 0:

so the excess temperature decays exponentially. Three consequences the paper uses:

• The hotter body cools faster. At 90 °C in a 25 °C room the body loses heat five times as fast as at 38 °C. This is why a cup of tea is drinkable in ten minutes but takes an hour to come down to room temperature.
• A plot of \ln(\theta-\theta_s) against t is a straight line of slope -K. That is the standard experimental verification of the law.
• The law is only a small-excess approximation to Stefan's law. For a body much hotter than its surroundings, radiation loss grows as T^4 and Newton's law fails.

Cooling correction. In a real mixture experiment the calorimeter loses heat to the room while the mixture settles. Newton's law supplies the fix: plot the temperature against time, and extrapolate the cooling curve backwards to the instant of mixing. The corrected final temperature is slightly higher than the highest reading, and the corrected initial rate is read from the same graph. This is why the experiment is stirred quickly, started close to room temperature, and finished fast.

9.4 Latent heat of fusion and vaporization; the triple point

Latent heat

While ice melts at 0 °C, its temperature does not rise, however hard you heat it. The heat is doing something else: it is pulling the molecules out of the rigid lattice and into the liquid. The energy goes into potential energy of the molecules, not into their kinetic energy — and temperature is a measure of the average kinetic energy. Only when the last crystal has melted does the thermometer start to climb again.

Change	Substance	Specific latent heat
Fusion (melting)	Ice at 0 °C	3.36\times10^5 J kg⁻¹ (80 cal g⁻¹)
Vaporization	Water at 100 °C	2.26\times10^6 J kg⁻¹ (540 cal g⁻¹)
Sublimation	Ice → vapour at 0 °C	2.83\times10^6 J kg⁻¹

Note how large L_v is: boiling 1 g of water takes about 5.4 times as much heat as heating it from 0 °C to 100 °C. That is why steam is such an effective heating medium, and why a burn from steam is far worse than a burn from boiling water: the steam first gives up 2260 J g⁻¹ of latent heat as it condenses on the skin, and only then does the resulting 100 °C water cool.

Evaporation versus boiling. Evaporation happens at any temperature, from the surface only, and cools the remaining liquid (the fastest molecules leave). Boiling happens at one fixed temperature for a given pressure, throughout the liquid, when the saturated vapour pressure equals the external pressure. That is why water boils at about 95 °C in Kathmandu (air pressure ≈ 0.88 atm) and at 71 °C on the summit of Sagarmatha — and why a pressure cooker, by raising the pressure, raises the boiling point and cooks dal faster.

The heating curve

The curve is the whole of calorimetry in one picture. To find the heat to take a mass m from one point to another, add up the straight sloped pieces (mc\Delta T) and the flat pieces (mL) that lie between them.

Doing the phase bookkeeping

An ice problem is not one equation but a decision tree. Work in this order and never skip step 2:

1. Heat available. How much heat would the warm bodies release if they cooled to 0 °C? (Include the calorimeter.)
2. Heat needed to melt all the ice. Q = mL_f. Compare it with step 1. - If step 1 < step 2, the final temperature is 0 °C and some ice survives. The answer is "0 °C, with m_{left} = m - Q_1/L_f of ice remaining". - If step 1 > step 2, all the ice melts; go on.
3. Leftover heat. Subtract the melting heat from the available heat. Use the remainder to warm the whole final mass — original water plus calorimeter plus melted ice — and solve for \Delta T.

The triple point

Why the paper cares about it:

• It is a single, reproducible fixed point — one temperature and one pressure — so it is far better as a thermometer calibration than the melting point, which depends on pressure. The kelvin was defined from the triple point of water from 1954 until the 2019 redefinition of the SI.
• Melting point ≠ triple point. Ice melts at 273.15 K (0.00 °C) under 1 atm, but at 273.16 K under 611 Pa. The two differ by 0.01 K because the melting point depends on pressure.
• The triple point pressure is below atmospheric, so ice does not sublime at 1 atm; it melts. Sublimation of ice is common in the dry, thin air of the high Himalayas, where the vapour pressure is very low.

The same idea applies to carbon dioxide, whose triple point is at 5.1 atm — above atmospheric pressure. That is why "dry ice" does not melt at room pressure: it sublimes straight to gas.

Formula and fact sheet

Topic	Formula or fact
Linear expansion	\Delta L = L\alpha\Delta T; L' = L(1+\alpha\Delta T)
Area, volume	\Delta A = A\beta\Delta T, \Delta V = V\gamma\Delta T
The relation	\beta = 2\alpha, \gamma = 3\alpha (isotropic solid, \alpha\Delta T\ll1)
Units	all three in K⁻¹; α (solids) ~10⁻⁵, γ (liquids) ~10⁻⁴, γ (gases) ~10⁻³
Hole in a plate	grows on heating, exactly as a disc of the same material would
Bimetallic strip	bends with the larger-α metal on the outside when heated; towards the smaller-α metal
Thermal stress	\sigma = Y\alpha\Delta T (compressive on heating), F = YA\alpha\Delta T
Pendulum clock	\dfrac{\Delta T_p}{T_p} = \dfrac{1}{2}\alpha\Delta\theta; loses time when hot
Time lost per day	\dfrac{1}{2}\alpha\Delta\theta\times86400 s
Real vs apparent	\gamma_r = \gamma_a + \gamma_v, with \gamma_v = 3\alpha_{vessel}
Overflow	\Delta V = V(\gamma_r - \gamma_v)\Delta T
Density on heating	\rho = \rho_0(1+\gamma\Delta T)^{-1}\approx\rho_0(1-\gamma\Delta T)
Water anomaly	maximum density 1000 kg m⁻³ at 4 °C; ice 917 kg m⁻³
Heat capacity	C = Q/\Delta T (J K⁻¹); C = mc
Specific heat	Q = mc\Delta T (J kg⁻¹ K⁻¹)
Water equivalent	W = m_cc_c/c_w; heat capacity of calorimeter = Wc_w
Method of mixtures	m_1c_1(\theta_1-\theta) = (m_2c_w + Wc_w)(\theta-\theta_2)
Latent heat	Q = mL (J kg⁻¹); L_f = 3.36\times10^5, L_v = 2.26\times10^6 J kg⁻¹
Sublimation	L_{sub}\approx L_f + L_v (with L_v at 0 °C) = 2.83\times10^6 J kg⁻¹
Newton's cooling	\dfrac{dQ}{dt} = k(\theta-\theta_s); \theta-\theta_s = (\theta_0-\theta_s)e^{-Kt}
Cooling check	\ln(\theta-\theta_s) vs t is a straight line of slope -K
Triple point of water	273.16 K (0.01 °C), 611 Pa (4.58 mm Hg); all three phases coexist
Useful constants	1 cal = 4.2 J; c_w = 4200, c_{ice} = 2100, c_{steam} = 2010 J kg⁻¹ K⁻¹

Chapter summary

• \Delta L = L\alpha\Delta T, \Delta A = A\beta\Delta T, \Delta V = V\gamma\Delta T, with \beta = 2\alpha and \gamma = 3\alpha for an isotropic solid. Area takes 2\alpha, volume takes 3\alpha — never the other way round.
• A hole or a cavity in a heated plate grows by the same fraction as the metal itself; nothing shrinks on heating except water below 4 °C.
• A clamped rod develops \sigma = Y\alpha\Delta T, compressive on heating. A pendulum clock runs slow when hot, losing \frac{1}{2}\alpha\Delta\theta of its time.
• A liquid's real expansion exceeds its apparent expansion by the vessel's: \gamma_r = \gamma_a + \gamma_v. The level in a warmed vessel first dips (the glass grows first) and then rises.
• Water is densest at 4 °C; that is why ice floats, ponds freeze from the top, and pipes burst in a hard frost.
• Q = mc\Delta T raises temperature; Q = mL changes phase at constant temperature. Never use both for the same stage.
• The calorimeter's water equivalent W = m_cc_c/c_w must be added to the water mass. Ignoring it makes the hot body's specific heat come out too low.
• In an ice problem, first compare the heat available with mL_f. If the ice cannot all melt, the final temperature is 0 °C with ice left over.
• Newton's law of cooling gives exponential cooling; equal temperature drops take longer and longer, and the hotter body cools faster.
• The triple point of water is 273.16 K at 611 Pa — the one state where ice, water and vapour coexist, and the reason the melting point (273.15 K at 1 atm) is not the same number.

Practice questions

Set A — Recall (23 questions)

1. For an isotropic solid, the coefficient of cubical expansion γ is related to the coefficient of linear expansion α byaγ = αbγ = 2αcγ = 3αdγ = α/3
2. The SI unit of the coefficient of linear expansion isambm K⁻¹cJ K⁻¹dK⁻¹
3. A circular hole is cut in a metal plate. When the plate is heated, the diameter of the holeadecreasesbincreasescstays the samedincreases only if the hole is at the centre
4. A bimetallic strip of brass and iron is heated. Brass has the larger α. The strip bends withabrass on the outer (convex) sidebiron on the outer sidecbrass on the inner sidedno bending, since both expand
5. Water has its maximum density ata0 °Cb2 °Cc10 °Cd4 °C
6. The SI unit of specific heat capacity isaJ K⁻¹bJ kg⁻¹cJ kg⁻¹ K⁻¹dkg J⁻¹ K⁻¹
7. The water equivalent of a calorimeter isathe mass of water with the same heat capacity as the calorimeterbthe mass of the calorimetercthe heat capacity of the calorimeterdthe mass of water in the calorimeter
8. The specific latent heat of fusion of ice isa2.26 × 10⁶ J kg⁻¹b3.36 × 10⁵ J kg⁻¹c4.2 × 10³ J kg⁻¹d3.36 × 10⁴ J kg⁻¹
9. The specific latent heat of vaporization of water at 100 °C is abouta3.36 × 10⁵ J kg⁻¹b2.26 × 10⁵ J kg⁻¹c4.2 × 10³ J kg⁻¹d2.26 × 10⁶ J kg⁻¹
10. The triple point of water is ata273.15 K and 1 atmb273.16 K and 611 Pac373.15 K and 1 atmd273.16 K and 1 atm
11. At the triple point of a substanceaonly the solid and liquid phases coexistbonly the liquid and vapour phases coexistcall three phases coexist in equilibriumdthe substance has no latent heat
12. If γₐ is the apparent expansivity of a liquid and γᵥ that of its vessel, the real expansivity isaγₐ + γᵥbγₐ − γᵥcγₐγᵥdγᵥ − γₐ
13. For the same rise in temperature, the cubical expansion is greatest inasolidsbliquidscgasesdall the same
14. Newton's law of cooling states that the rate of loss of heat by a body is proportional toaits absolute temperaturebthe fourth power of its absolute temperaturecthe square of the excess temperaturedits excess temperature over the surroundings
15. The dimensional formula of the coefficient of linear expansion isa[K⁻¹]b[K]c[L K⁻¹]d[L⁻¹ K⁻¹]
16. The heat capacity of a body is the heat needed toaraise unit mass through 1 Kbraise its temperature by 1 Kcmelt it completelydraise it from 0 °C to 100 °C
17. When a liquid in a glass vessel is warmed gently, the level first falls and then rises becauseathe liquid contracts at firstbthe liquid evaporatescglass has a larger γ than the liquiddthe vessel expands at first, before the liquid's expansion overtakes it
18. The coefficient of superficial expansion β equalsaαb2αc3αdα/2
19. A rod of length L and coefficient α is heated through ΔT. Its increase in length isaLαΔTbLα/ΔTc2LαΔTd3LαΔT
20. A rod is clamped at both ends and then heated. The stress developed in it isazero, because it cannot expandbtensile, of magnitude YαΔTccompressive, of magnitude YαΔTdYΔT/α
21. The principle of calorimetry states thatathe final temperature is the mean of the initial temperaturesbheat lost by the hot body equals heat gained by the cold body if none escapescthe total heat of the system is zerodheat flows from the colder to the hotter body
22. Which of the following substances has the highest specific heat capacity?acopperbmercurycirondwater
23. Two bodies of the same mass need the same heat for the same temperature rise when they have the sameadensitybvolumecspecific heat capacitydthermal conductivity

Answers and explanations — Set A

1. (c) — length is one dimension, area two, volume three, so the coefficients go as 1 : 2 : 3. Option (b) is the area coefficient.

2. (d) — α is a fractional change of length per kelvin, so the metre cancels and K⁻¹ is left.

3. (b) — every linear dimension of the plate grows by the fraction αΔT, and the hole is one of them. A hole behaves like a disc of the same metal.

4. (a) — brass lengthens more, so it must form the longer (outer) arc. The strip curves towards the metal with the smaller α.

5. (d) — between 0 °C and 4 °C water contracts, and above 4 °C it expands; the turning point is the density maximum.

6. (c) — heat per unit mass per kelvin: J/(kg K).

7. (a) — the water equivalent is a mass of water whose heat capacity equals the calorimeter's. Option (c) is the heat capacity itself, in J K⁻¹.

8. (b) — 80 cal g⁻¹ × 4.2 = 336 J g⁻¹ = 3.36 × 10⁵ J kg⁻¹. The 10⁶ value in (a) is the latent heat of vaporization.

9. (d) — 540 cal g⁻¹ × 4.2 ≈ 2.26 × 10⁶ J kg⁻¹, about seven times the latent heat of fusion.

10. (b) — 273.16 K at 611 Pa. Option (a) is the ordinary melting point at 1 atm.

11. (c) — that is the definition: solid, liquid and vapour in equilibrium at one temperature and pressure.

12. (a) — the vessel's growth hides part of the liquid's growth, so the real coefficient is the larger one.

13. (c) — gases have γ ≈ 3.7 × 10⁻³ K⁻¹, some twenty times a liquid's and a few hundred times a solid's.

14. (d) — the driving difference is the body's temperature minus the surroundings'. Option (b) is Stefan's law, which Newton's law approximates for small excess.

15. (a) — (1/L)(ΔL/ΔT) = [L]/[L][K] = [K⁻¹].

16. (b) — heat capacity is per kelvin for the whole body; per unit mass is the specific heat, option (a).

17. (d) — the glass in contact with the warm liquid heats first and its capacity grows before the bulk liquid has expanded much.

18. (b) — area is two-dimensional, so β = 2α. Option (c) belongs to volume.

19. (a) — the definition of α rearranged: ΔL = LαΔT.

20. (c) — the rod would have grown, so it is compressed; the strain αΔT multiplied by Y gives the stress.

21. (b) — conservation of energy for an isolated system. The final temperature is a weighted mean, not a simple mean, unless the masses and specific heats are equal.

22. (d) — water's 4200 J kg⁻¹ K⁻¹ is about ten times copper's and thirty times mercury's.

23. (c) — Q = mcΔT, so equal m, ΔT and Q requires equal c.

Set B — Understanding (15 questions)

1. A pendulum clock keeps correct time at 20 °C. On a day when the temperature is 35 °C, the clockagains time, because the pendulum shortensbloses time, because the pendulum lengthensckeeps correct time, because g is unchangeddloses time, because g decreases
2. Two rods of the same length are made of metals with α in the ratio 1 : 2. Heated through the same ΔT, the ratio of their increases in length isa2 : 1b1 : 4c4 : 1d1 : 2
3. A cube is heated through ΔT. The fractional changes in its length, surface area and volume are in the ratioa1 : 2 : 3b1 : 3 : 2c3 : 2 : 1d1 : 1 : 1
4. A bimetallic strip of brass and iron is cooled below room temperature. Compared with its behaviour on heating, it bendsathe same waybthe opposite way, with brass on the inner sidecthe opposite way, with brass on the outer sidednot at all
5. The apparent expansion of a liquid in a vessel is found to be zero. This means thatathe liquid and the vessel expand equallybthe liquid does not expandcthe vessel does not expanddthe liquid contracts
6. Ice at 0 °C cools a drink better than the same mass of water at 0 °C becauseaice is colder than waterbice has a lower specific heatcice absorbs latent heat as it meltsdice has a higher density
7. Steam at 100 °C causes a worse burn than water at 100 °C because steamais at a higher temperaturebhas a higher specific heatcreleases latent heat when it condensesdis a gas and spreads faster
8. In a mixture experiment the water equivalent of the calorimeter is ignored. The specific heat calculated for the hot solid will beatoo highbtoo lowcunaffecteddexactly zero
9. A solid is heated and its volume increases by 0.3 %. Its densityaincreases by 0.3 %bdecreases by 0.3 %cdecreases by 0.1 %dis unchanged
10. The rate of cooling of a hot body is greatestawhen it is hottestbwhen it is near room temperaturecat a steady rate throughoutdonly at the moment it reaches room temperature
11. The steel rod of a pendulum clock is replaced by an invar rod of the same length. The clock is nowaalmost unaffected by temperature changesbmore affected by temperature changescslower at every temperaturedfaster at every temperature
12. Two solid spheres of the same metal have radii in the ratio 1 : 2. Heated through the same ΔT, the ratio of the increases in their volumes isa1 : 2b1 : 4c1 : 6d1 : 8
13. Water at 0 °C is warmed slowly. Its volumeaincreases steadilybdecreases steadilycdecreases until 4 °C, then increasesdincreases until 4 °C, then decreases
14. The temperature of a body does not change while it melts, although heat is supplied. The heat suppliedais lost to the surroundingsbraises the kinetic energy of the moleculescis used only to expand the bodydraises the potential energy of the molecules
15. Equal masses of water and kerosene (c = 2100 J kg⁻¹ K⁻¹) absorb equal amounts of heat. The rise in temperature of the kerosene isahalf that of the waterbtwice that of the watercthe same as that of the waterdfour times that of the water

Answers and explanations — Set B

1. (b) — a longer pendulum has a longer period T = 2π√(L/g), so each swing takes more time and the clock falls behind. g does not change measurably with the weather.

2. (d) — ΔL = LαΔT with L and ΔT the same, so ΔL ∝ α: the ratio is 1 : 2.

3. (a) — ΔL/L = αΔT, ΔA/A = 2αΔT, ΔV/V = 3αΔT, so the fractional changes are in the ratio of the dimensions, 1 : 2 : 3.

4. (b) — on cooling, brass contracts more than iron, so brass becomes the shorter (inner) surface. The bend reverses direction.

5. (a) — γₐ = γᵣ − γᵥ = 0 means γᵣ = γᵥ. The liquid does expand; the vessel expands by exactly as much, so nothing seems to change.

6. (c) — melting takes 336 J g⁻¹ of latent heat from the drink before the ice water can warm at all. The temperature of both is 0 °C, so (a) is wrong.

7. (c) — condensing steam gives up 2260 J g⁻¹ before the 100 °C water even starts to cool. Both are at the same temperature.

8. (b) — the heat actually gained by the cold side is larger than m_w c_w Δθ; ignoring W makes the numerator too small, so c comes out too low.

9. (b) — ρ = m/V, so a fractional rise γΔT in volume gives the same fractional fall in density.

10. (a) — the rate is proportional to the excess temperature, which is largest at the start. That is why the cooling curve is steepest at the beginning.

11. (a) — invar's α is about 1.2 × 10⁻⁶ K⁻¹, ten times smaller than steel's, so the length hardly changes and the clock keeps time in all weathers.

12. (d) — volumes go as the cube of the radius, 1 : 8, and ΔV = VγΔT with the same γ and ΔT.

13. (c) — water contracts from 0 °C to 4 °C (the anomalous region) and expands above 4 °C, so the volume is least at 4 °C.

14. (d) — the heat breaks the bonds of the lattice, increasing the molecules' potential energy. The average kinetic energy, and therefore the temperature, stays the same.

15. (b) — Δθ = Q/(mc), so with Q and m equal, Δθ ∝ 1/c: kerosene's c is half of water's, so its rise is twice as large.

Set C — Application (9 questions)

1. A steel rail is 1.0 m long at 20 °C (α = 1.2 × 10⁻⁵ K⁻¹). Its increase in length at 70 °C isa0.6 mmb0.06 mmc6 mmd0.3 mm
2. A pendulum clock with a steel rod (α = 2 × 10⁻⁵ K⁻¹) keeps correct time at 20 °C. The time it loses in a day at 35 °C is abouta26 sb6.5 sc1.3 sd13 s
3. A glass vessel (γ = 2.4 × 10⁻⁵ K⁻¹) is filled with a liquid whose apparent expansivity is 1.6 × 10⁻⁴ K⁻¹. The real expansivity of the liquid isa1.36 × 10⁻⁴ K⁻¹b1.6 × 10⁻⁴ K⁻¹c1.84 × 10⁻⁴ K⁻¹d4.0 × 10⁻⁵ K⁻¹
4. 100 g of water at 60 °C is mixed with 50 g of water at 20 °C in a calorimeter of water equivalent 10 g. The final temperature isa40 °Cb45 °Cc46.7 °Cd50 °C
5. 20 g of ice at 0 °C is dropped into 100 g of water at 40 °C. Neglecting the calorimeter, the final temperature is (L_f = 3.36\times10^5 J kg⁻¹)a0 °Cb10 °Cc20 °Cd24 °C
6. The heat needed to convert 10 g of ice at −10 °C into steam at 100 °C is (c_{ice} = 2100, c_w = 4200, L_f = 3.36\times10^5, L_v = 2.26\times10^6, all SI)a7.8 × 10³ Jb3.04 × 10⁴ Jc2.6 × 10⁴ Jd3.0 × 10⁵ J
7. A steel rod (Y = 2 × 10¹¹ Pa, α = 1.2 × 10⁻⁵ K⁻¹) is clamped at both ends and heated through 100 K. The stress set up isa1.2 × 10⁸ Pab2.4 × 10⁶ Pac4.8 × 10⁸ Pad2.4 × 10⁸ Pa
8. A body cools from 60 °C to 50 °C in 10 minutes in surroundings at 25 °C. The time it takes to cool from 50 °C to 45 °C is abouta6.7 minb5.0 minc8.0 mind10 min
9. The mass of ice at 0 °C needed to cool 200 g of water from 50 °C to 0 °C is (L_f = 3.36\times10^5 J kg⁻¹)a100 gb125 gc150 gd200 g

Answers and explanations — Set C

1. (a) — ΔL = 1.0 × 1.2 × 10⁻⁵ × 50 = 6.0 × 10⁻⁴ m = 0.6 mm. Option (b) uses ΔT = 5 K.

2. (d) — fractional loss ½αΔθ = ½(2 × 10⁻⁵)(15) = 1.5 × 10⁻⁴; per day 1.5 × 10⁻⁴ × 86400 ≈ 13 s. Option (a) forgets the ½.

3. (c) — γᵣ = γₐ + γᵥ = 1.6 × 10⁻⁴ + 0.24 × 10⁻⁴ = 1.84 × 10⁻⁴ K⁻¹. Option (a) subtracts instead of adding.

4. (b) — (100 × 60 + 60 × 20)/(100 + 60) = 7200/160 = 45 °C. The 10 g water equivalent must be added to the cold water, not to the hot.

5. (c) — heat available = 0.100 × 4200 × 40 = 16 800 J; melting 20 g needs 6720 J, so all the ice melts. Remaining 10 080 J warms 120 g: Δθ = 10 080/(0.120 × 4200) = 20 K, final 20 °C.

6. (b) — 210 + 3360 + 4200 + 22 600 = 30 370 J ≈ 3.04 × 10⁴ J. Option (a) omits the boiling stage; (c) counts only the melting and boiling stages.

7. (d) — σ = YαΔT = 2 × 10¹¹ × 1.2 × 10⁻⁵ × 100 = 2.4 × 10⁸ Pa, compressive.

8. (a) — mean excess 30 K gives 10 K in 10 min; for the next 5 K the mean excess is 22.5 K, so the rate is 0.75 K min⁻¹ and the time is 5/0.75 ≈ 6.7 min. The exponential law gives 6.6 min.

9. (b) — heat to remove = 0.200 × 4200 × 50 = 42 000 J; m = 42 000/(3.36 × 10⁵) = 0.125 kg = 125 g.