Part 2 · Chapter 18

Temperature, Heat, and the First Law of Thermodynamics

Why a lid loosens under hot water, why a lake freezes from the top down, and why boiling 1 kg of water needs five times the energy of warming it — all from one bookkeeping rule for energy

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 60 min
i What you'll learn
  • That the zeroth law makes temperature meaningful, and that the SI Kelvin scale is anchored at the triple point of water, \(T_{3} = 273.16\,\mathrm{K}\).
  • How the everyday scales relate to it: \(T_{C} = T - 273.15^{\circ}\) and \(T_{F} = \tfrac{9}{5}T_{C} + 32^{\circ}\).
  • Thermal expansion — linear \(\Delta L = L\alpha\,\Delta T\) and volume \(\Delta V = V\beta\,\Delta T\) with \(\beta = 3\alpha\).
  • That heat \(Q\) is energy in transit, with \(Q = cm\,\Delta T\) for a temperature change and \(Q = Lm\) for a phase change.
  • The work a gas does, \(W = \int p\,dV\), and the first law \(\Delta E_{\text{int}} = Q - W\), whose special cases (adiabatic, constant-volume, cyclical, free expansion) follow by setting one term to zero.
  • The three transfer mechanisms — conduction \(P_{\text{cond}} = kA\,\Delta T/L\), convection, and radiation \(P_{\text{rad}} = \sigma\varepsilon A T^{4}\).
Section 18-1

What Is Physics?

One of the great branches of physics and engineering is thermodynamics — the study of the thermal (internal) energy of systems and how it moves around. You have been building an intuition for it since childhood: you are careful with a hot stove, you keep milk in the cold part of the fridge, and you know how to dress against wind chill. The central concept underneath all of it is temperature.

The reach of thermodynamics is enormous. Automotive engineers manage the heat of a racing engine; food engineers worry about microwaving a pizza evenly and flash-freezing a meal; geologists track the thermal energy moving through an El Niño or a melting ice sheet; medical engineers ask whether a patient's temperature signals a harmless infection or something worse. Every one of these problems begins with two questions: what is temperature, and how do we measure it?

Section 18-2

Temperature and the Zeroth Law

Temperature is one of the seven SI base quantities. Physicists measure it on the Kelvin scale, in units called kelvins. Temperature has no known upper limit — the universe just after the Big Bang was about \(10^{39}\,\mathrm{K}\) — but it does have a lower limit, absolute zero, taken as the zero of the Kelvin scale. Room temperature is about 290 K above it.

Many properties of a body change as its temperature changes: a liquid's volume, a metal rod's length, a wire's electrical resistance, the pressure of a confined gas. Any one of these can be the working part of a thermoscope — a device whose readout rises when heated and falls when cooled. But a bare thermoscope is not yet a thermometer; its numbers have no agreed meaning until we calibrate it. To do that, we need one experimental fact.

Put a thermoscope (call it body T) in intimate contact with body A inside an insulating box and wait. Its reading drifts and then settles: bodies T and A are now in thermal equilibrium. Do the same with body B and find the same final reading. Now bring A and B into contact directly — and experiment shows they need no adjustment at all. They were already in equilibrium. That fact is the zeroth law of thermodynamics.

🌡️
The Zeroth Law of Thermodynamics
If A and B are each in thermal equilibrium with a third body T, then A and B are in equilibrium with each other.

In plain words: every body has a property called temperature; when two bodies are in thermal equilibrium their temperatures are equal, and vice versa. This is exactly what lets a thermometer work — we never bring two coffee cups into contact to compare them, we just read each with the same instrument.

Why "zeroth"? The law was recognized in the 1930s, long after the first and second laws had already been named. But because the very concept of temperature that those two laws rely on is established by this one, it had to come logically first — so it was given the number that comes before one.
Section 18-3

Measuring Temperature: The Kelvin Scale

To build a scale we choose one reproducible thermal event, assign it a number by agreement, and calibrate against it. We could use water's freezing or boiling point, but both depend on pressure. Instead we use the triple point of water — the single pressure–temperature combination at which ice, liquid water, and water vapor coexist in equilibrium. It is sharp and reproducible, and by international agreement it is assigned:

Triple-point temperature (the anchor of the Kelvin scale)
\[ T_{3} = 273.16\,\mathrm{K} \]
This single choice also fixes the size of the kelvin as \(1/273.16\) of the interval from absolute zero to the triple point. Note: we write "300 K" and say "300 kelvins" — no degree sign, and never "degrees Kelvin."

The standard instrument that realizes this scale is the constant-volume gas thermometer: a gas-filled bulb connected to a mercury manometer, with the gas volume held fixed. The temperature is taken proportional to the gas pressure, so comparing the pressure \(p\) at the unknown temperature with the pressure \(p_{3}\) at the triple point gives:

Temperature from a constant-volume gas thermometer
\[ T = (273.16\,\mathrm{K})\left(\lim_{\text{gas}\to 0}\frac{p}{p_{3}}\right) \]
Different gases give slightly different readings, but as the amount of gas in the bulb shrinks toward zero, all gases converge to one value — the ideal-gas temperature. That limit is what the formula instructs us to extrapolate to.
Section 18-4

The Celsius and Fahrenheit Scales

The Kelvin scale is the scale of basic science, but day-to-day life uses the Celsius scale (and, in a few countries, Fahrenheit). A Celsius degree is exactly the same size as a kelvin; only the zero is shifted to a more convenient place:

Kelvin → Celsius, and Celsius → Fahrenheit
\[ T_{C} = T - 273.15^{\circ} \qquad\qquad T_{F} = \tfrac{9}{5}T_{C} + 32^{\circ} \]
A Fahrenheit degree is smaller than a Celsius degree (\(5\,\mathrm{C^{\circ}} = 9\,\mathrm{F^{\circ}}\)) and the zero is different again. Anchor points to memorize: water freezes at \(0^{\circ}\mathrm{C} = 32^{\circ}\mathrm{F}\), boils at \(100^{\circ}\mathrm{C} = 212^{\circ}\mathrm{F}\), and the two scales coincide at \(-40^{\circ}\).
Table 18-1 · Some corresponding temperatures
Reference point°C°F
Boiling point of water100212
Normal body temperature37.098.6
Comfort level2068
Freezing point of water032
Zero of Fahrenheit scale−180
Scales coincide−40−40
Degrees vs degree-differences. Write \(0^{\circ}\mathrm{C} = 32^{\circ}\mathrm{F}\) for a temperature, but \(5\,\mathrm{C^{\circ}} = 9\,\mathrm{F^{\circ}}\) for a temperature difference (the degree symbol moves after the letter). Mixing these up is the single most common scale-conversion slip.
Section 18-5

Thermal Expansion

Hold a stuck metal jar lid under hot water and it loosens, because the metal expands more than the glass beneath it. As atoms gain energy they sit a little farther apart against the spring-like bonds that hold a solid together — so the whole object swells. Engineers must anticipate this everywhere: expansion slots in bridges, dental fillings matched to tooth enamel, the Concorde's fuselage stretching about 12.5 cm in supersonic flight.

For a rod of length \(L\) heated by \(\Delta T\), the length grows by an amount set by the coefficient of linear expansion \(\alpha\):

Linear and volume expansion
\[ \Delta L = L\,\alpha\,\Delta T \qquad\qquad \Delta V = V\,\beta\,\Delta T \qquad\qquad \beta = 3\alpha \]
Thermal expansion is like a three-dimensional photographic enlargement: every length scales by the same factor — including the diameter of a hole, which grows as if it were filled with material. For liquids, only volume expansion is meaningful, and \(\beta = 3\alpha\) follows from expanding a cube in all three directions.
Table 18-2 · Some coefficients of linear expansion, \(\alpha\;(10^{-6}/\mathrm{C^{\circ}})\)
SubstanceαSubstanceα
Ice (0 °C)51Steel11
Lead29Glass (ordinary)9
Aluminum23Glass (Pyrex)3.2
Brass19Diamond1.2
Copper17Invar0.7
Concrete12Fused quartz0.5
Water's strange exception. Above about 4 °C water expands as it warms, as expected. But between 0 and 4 °C it contracts as it warms, so its density peaks near 4 °C. This is why lakes freeze from the top down: the coldest surface water is the lightest, so it stays on top and freezes there, leaving liquid water — and aquatic life — protected below.
Section 18-6

Temperature and Heat

Leave a cold cola can on the table and it warms until it reaches room temperature; leave hot coffee and it cools to the same point. Call the cola or coffee the system (temperature \(T_{S}\)) and the kitchen its environment (temperature \(T_{E}\)). Whenever \(T_{S} \neq T_{E}\), energy flows until the two match. That transferred energy — driven purely by a temperature difference — is heat, symbol \(Q\).

🔥
Definition of heat
Heat is energy transferred between a system and its environment because of a temperature difference between them.

\(Q > 0\) when energy flows into the system (absorbed); \(Q < 0\) when it flows out (released); \(Q = 0\) at equilibrium. Like work, heat is not something a body "contains" — it only describes energy in transit. It is meaningless to say a system "has 450 J of heat," just as a bank statement reports transfers, not a sum the account "is made of."

Before heat was understood as transferred energy, it was measured by its power to warm water: the calorie was the heat that raises 1 g of water by 1 C° (14.5 to 15.5 °C). Since 1948 the SI unit is the joule, and the conversions are fixed:

Units of heat
\[ 1\,\mathrm{cal} = 3.968 \times 10^{-3}\,\mathrm{Btu} = 4.1868\,\mathrm{J} \]
The nutritional "Calorie" (capital C) is actually a kilocalorie — 1000 of these calories.
Section 18-7

The Absorption of Heat by Solids and Liquids

The heat capacity \(C\) of an object links the heat it absorbs to its temperature change, \(Q = C\,\Delta T\). Because two objects of the same material have heat capacities in proportion to their masses, it is more useful to quote a specific heat \(c\) — heat capacity per unit mass — which is a property of the material itself:

Heat to change temperature
\[ Q = C\,\Delta T \qquad\qquad Q = cm\,\Delta T = cm(T_{f} - T_{i}) \]
Water's specific heat is unusually large: \(c = 1\,\mathrm{cal/g\cdot C^{\circ}} = 4187\,\mathrm{J/kg\cdot K}\). This is why coastal climates are mild and why water is such a good coolant — it soaks up a lot of energy for a small temperature rise.
Table 18-3 · Specific heats at room temperature, \(c\;(\mathrm{J/kg\cdot K})\)
SubstancecSubstancec
Lead128Glass840
Copper386Aluminum900
Brass380Ice (−10 °C)2220
Granite790Ethyl alcohol2430
Seawater3900Water4187

Sometimes added heat does not raise the temperature at all — instead it drives a phase change (melting, freezing, boiling, condensing), going entirely into rearranging the molecular bonds. The energy per unit mass for a complete phase change is the heat of transformation \(L\):

🧊
Heat for a phase change
Q = Lm

For water, the heat of fusion (solid ↔ liquid) is \(L_{F} = 333\,\mathrm{kJ/kg}\) and the heat of vaporization (liquid ↔ gas) is \(L_{V} = 2256\,\mathrm{kJ/kg}\). These are large: melting ice at 0 °C takes far more energy than warming the same mass of liquid water by many degrees, which is exactly why ice is so effective at keeping a drink cold.

Section 18-8

A Closer Look at Heat and Work

Take a gas confined in a cylinder with a movable, loaded piston, sitting on a thermal reservoir whose temperature you can dial. Remove a little of the load and the gas pushes the piston up through a tiny displacement \(ds\) with force \(pA\). The small work done by the gas is \(dW = pA\,ds = p\,dV\), so over a finite expansion:

Work done by a gas
\[ W = \int dW = \int_{V_{i}}^{V_{f}} p\,dV \]
On a p–V diagram, this work is the area under the curve traced from the initial state \(i\) to the final state \(f\). Expansion (volume increasing) gives positive work by the gas; compression gives negative work.

Crucially, a gas can travel from the same state \(i\) to the same state \(f\) along infinitely many paths, and the area under the curve — hence the work \(W\) — differs for each. The same is true of the heat \(Q\). Heat and work are path-dependent: neither is a property of the state alone.

Why p–V area matters. Run a closed loop on a p–V diagram — out along one path, back along another — and the gas does positive work going out and negative work coming back. The net work for the cycle is the enclosed area, positive for a clockwise loop. This is the seed of every heat engine in Chapter 20.
Section 18-9

The First Law of Thermodynamics

Here is the surprise. Although \(Q\) and \(W\) separately depend on the path, experiment shows that the combination \(Q - W\) is the same for every path between two states. It depends only on the endpoints — so it must be the change in an intrinsic property of the system, the internal energy \(E_{\text{int}}\):

⚖️
The First Law of Thermodynamics
ΔE_int = Q − W  ·  dE_int = dQ − dW

This is energy conservation extended to systems that exchange energy with their surroundings. Internal energy rises when heat is added to the system or when work is done on it; it falls when the system loses heat or does work on its surroundings. In terms of work done on the system, \(\Delta E_{\text{int}} = Q + W_{\text{on}}\).

Mind the sign convention. From this chapter onward, \(W\) means the work done by the system (the gas lifting the piston). That is why it carries a minus sign in \(\Delta E_{\text{int}} = Q - W\): work done by the gas drains its internal energy. Earlier chapters used \(W\) for work done on a body — same physics, opposite sign.
Section 18-10

Some Special Cases of the First Law

The power of \(\Delta E_{\text{int}} = Q - W\) is that whole families of processes appear simply by setting one term to zero. Four cases recur constantly.

Four special cases of ΔE_int = Q − W
\[\begin{aligned} \text{Adiabatic } (Q=0): &\quad \Delta E_{\text{int}} = -W \\ \text{Constant volume } (W=0): &\quad \Delta E_{\text{int}} = Q \\ \text{Cyclical } (\Delta E_{\text{int}}=0): &\quad Q = W \\ \text{Free expansion } (Q=W=0): &\quad \Delta E_{\text{int}} = 0 \end{aligned}\]
Adiabatic — no heat crosses the boundary (well-insulated or very fast), so work comes straight out of internal energy. Constant volume — the gas can do no work, so all heat becomes internal energy. Cyclical — the system returns to its start, so net heat equals net work (a heat engine). Free expansion — gas rushing into a vacuum does no work and exchanges no heat, so its internal energy is unchanged.
Table 18-5 · The first law: four special cases
ProcessRestrictionConsequence
Adiabatic\(Q = 0\)\(\Delta E_{\text{int}} = -W\)
Constant volume\(W = 0\)\(\Delta E_{\text{int}} = Q\)
Closed cycle\(\Delta E_{\text{int}} = 0\)\(Q = W\)
Free expansion\(Q = W = 0\)\(\Delta E_{\text{int}} = 0\)
A free expansion can't be drawn. Unlike the others, a free expansion happens too suddenly to keep the gas in equilibrium — its pressure isn't even uniform mid-rush. We can mark the initial and final states on a p–V diagram, but not the path between them.
Section 18-11

Heat Transfer Mechanisms

We have said energy moves as heat — but how? There are three mechanisms: conduction, convection, and radiation.

Conduction. Leave a metal poker in a fire and its handle grows hot. Energetic atoms at the hot end jostle their neighbors, passing vibrational energy down the metal collision by collision. For a slab of area \(A\) and thickness \(L\) with its faces held at \(T_{H}\) and \(T_{C}\), the steady conduction rate is:

Conduction rate, and the R-value
\[ P_{\text{cond}} = \frac{Q}{t} = kA\,\frac{T_{H} - T_{C}}{L} \qquad\qquad R = \frac{L}{k} \]
\(k\) is the thermal conductivity: high for metals (copper ≈ 401 W/m·K), low for insulators (air ≈ 0.026, polyurethane foam ≈ 0.024). The R-value \(L/k\) describes a slab of a given thickness — a high R-value means a good insulator. For layers in series, \(P_{\text{cond}} = A(T_{H}-T_{C})/\sum(L/k)\).

Convection. Watch a candle flame and you watch convection: a fluid touching a hot object warms, expands, becomes less dense, and floats upward as buoyancy lifts it, while cooler fluid sinks to take its place. Convection drives weather and ocean currents, keeps gliders and birds aloft on thermals, and carries energy from the Sun's core to its surface.

Radiation. The third route needs no medium at all — energy travels as electromagnetic waves, which is how the Sun's warmth crosses empty space to reach you. Every object above absolute zero radiates:

Thermal radiation: emitted, and net exchanged
\[ P_{\text{rad}} = \sigma\varepsilon A T^{4} \qquad\qquad P_{\text{net}} = \sigma\varepsilon A\,(T_{\text{env}}^{4} - T^{4}) \]
\(\sigma = 5.6704 \times 10^{-8}\,\mathrm{W/m^{2}\cdot K^{4}}\) is the Stefan–Boltzmann constant; \(\varepsilon\) is the emissivity (0 to 1; an ideal blackbody has \(\varepsilon = 1\)). The temperature must be in kelvins, and the steep \(T^{4}\) dependence means a modest rise in temperature radiates dramatically more energy. \(P_{\text{net}} > 0\) means the object is gaining energy from its surroundings.
Worked Examples

Putting It to Work

1 Converting between two temperature scales

Problem. An old notebook describes a scale Z on which water boils at \(65.0^{\circ}\mathrm{Z}\) and freezes at \(-14.0^{\circ}\mathrm{Z}\). To what Fahrenheit temperature does \(T = -98.0^{\circ}\mathrm{Z}\) correspond? (Assume Z is linear.)

Solution. Compare degree sizes between matching benchmarks, then convert the offset from a known point. On Z the boiling–freezing span is \(79.0\,\mathrm{Z^{\circ}}\); on Fahrenheit it is \(180\,\mathrm{F^{\circ}}\). The target is below freezing by \(-14.0 - (-98.0) = 84.0\,\mathrm{Z^{\circ}}\).

Scale the offset, then subtract from the freezing point (32 °F)
\[ (84.0\,\mathrm{Z^{\circ}})\frac{180\,\mathrm{F^{\circ}}}{79.0\,\mathrm{Z^{\circ}}} = 191\,\mathrm{F^{\circ}} \quad\Rightarrow\quad T = 32.0^{\circ}\mathrm{F} - 191\,\mathrm{F^{\circ}} = -159^{\circ}\mathrm{F} \]

The trick is always to convert a difference from a benchmark, never to plug raw scale readings into a formula meant for Celsius and Fahrenheit alone.

2 Volume expansion of diesel fuel

Problem. A trucker loads \(37\,000\,\mathrm{L}\) of diesel on a hot day in Las Vegas, then delivers it where the temperature is \(23.0\,\mathrm{K}\) lower. How many liters arrive? (For diesel, \(\beta = 9.50 \times 10^{-4}/\mathrm{C^{\circ}}\).)

Solution. The fuel volume tracks temperature through \(\Delta V = V\beta\,\Delta T\).

ΔV = Vβ ΔT, then V_delivered = V + ΔV
\[ \Delta V = (37\,000)(9.50 \times 10^{-4})(-23.0) \approx -808\,\mathrm{L} \quad\Rightarrow\quad V_{\text{del}} = 37\,000 - 808 \approx 36\,190\,\mathrm{L} \]

About 808 L "vanish" — not lost, just contracted. The steel tank's expansion is irrelevant here; what matters is the fuel. (And it raises a fair question: who paid for the missing diesel?)

3 A hot copper slug reaching equilibrium in water

Problem. A copper slug (\(m_{c} = 75\,\mathrm{g}\), \(c_{c} = 0.0923\,\mathrm{cal/g\cdot K}\)) heated to \(312^{\circ}\mathrm{C}\) is dropped into \(m_{w} = 220\,\mathrm{g}\) of water at \(12^{\circ}\mathrm{C}\) in a beaker of heat capacity \(C_{b} = 45\,\mathrm{cal/K}\). Find the equilibrium temperature \(T_{f}\).

Solution. The system is isolated, so the three heat transfers sum to zero: \(Q_{w} + Q_{b} + Q_{c} = 0\). Solve for \(T_{f}\) (Celsius is fine — only differences appear).

c_w m_w(T_f − T_i) + C_b(T_f − T_i) + c_c m_c(T_f − T) = 0
\[ T_{f} = \frac{c_{c}m_{c}T + C_{b}T_{i} + c_{w}m_{w}T_{i}}{c_{w}m_{w} + C_{b} + c_{c}m_{c}} = \frac{5339.8\,\mathrm{cal}}{271.9\,\mathrm{cal/C^{\circ}}} \approx 19.6^{\circ}\mathrm{C} \approx 20^{\circ}\mathrm{C} \]

The slug loses about 2020 cal; the water gains about 1670 cal and the beaker about 342 cal — and the three indeed cancel. Water's large specific heat is why 220 g of it barely warms despite a slug arriving 300° hotter.

4 Heat to change both temperature and state

Problem. How much heat takes \(m = 720\,\mathrm{g}\) of ice at \(-10^{\circ}\mathrm{C}\) all the way to liquid water at \(15^{\circ}\mathrm{C}\)? Use \(c_{\text{ice}} = 2220\), \(c_{\text{liq}} = 4187\,\mathrm{J/kg\cdot K}\) and \(L_{F} = 333\,\mathrm{kJ/kg}\).

Solution. Three steps in sequence — warm the ice, melt it, warm the meltwater — because temperature and phase cannot change at the same time.

Q_tot = Q_warm-ice + Q_melt + Q_warm-water
\[\begin{aligned} Q_{1} &= c_{\text{ice}}m\,(10) = 15.98\,\mathrm{kJ} \\ Q_{2} &= L_{F}m = (333)(0.720) = 239.8\,\mathrm{kJ} \\ Q_{3} &= c_{\text{liq}}m\,(15) = 45.22\,\mathrm{kJ} \\[2pt] Q_{\text{tot}} &= 15.98 + 239.8 + 45.22 \approx 300\,\mathrm{kJ} \end{aligned}\]

Melting dominates — it takes far more energy than either warming step. In fact, supplying only 210 kJ would warm the ice to 0 °C (15.98 kJ) and then melt just part of it: the remaining 194 kJ melts \(m = 194/333 \approx 0.58\,\mathrm{kg}\), leaving 580 g of water and 140 g of ice, both at 0 °C.

5 First law for boiling water

Problem. Boil \(1.00\,\mathrm{kg}\) of water at \(100^{\circ}\mathrm{C}\) into steam at constant atmospheric pressure (\(1.01 \times 10^{5}\,\mathrm{Pa}\)). The volume jumps from \(1.00 \times 10^{-3}\,\mathrm{m^{3}}\) to \(1.671\,\mathrm{m^{3}}\). Find (a) the work done by the system, (b) the heat absorbed, and (c) the change in internal energy.

Solution. At constant pressure, \(W = p\,\Delta V\); the heat is a pure phase change, \(Q = L_{V}m\); then apply the first law.

W = p ΔV ; Q = L_V m ; ΔE_int = Q − W
\[\begin{aligned} W &= (1.01 \times 10^{5})(1.671 - 1.00\times10^{-3}) \approx 169\,\mathrm{kJ} \\ Q &= L_{V}m = (2256\,\mathrm{kJ/kg})(1.00\,\mathrm{kg}) \approx 2260\,\mathrm{kJ} \\ \Delta E_{\text{int}} &= Q - W = 2256 - 169 \approx 2090\,\mathrm{kJ} = 2.09\,\mathrm{MJ} \end{aligned}\]

Only about 7.5% of the heat goes into pushing back the atmosphere; the rest — over 2 MJ — goes into prying apart the strongly attracting water molecules. That stored bond energy is exactly what scalds you when steam condenses on skin.

6 Conduction through a window pane

Problem. A glass window of area \(A = 1.5\,\mathrm{m^{2}}\) and thickness \(L = 3.0\,\mathrm{mm}\) (\(k = 1.0\,\mathrm{W/m\cdot K}\)) has its inner face at \(20^{\circ}\mathrm{C}\) and outer face at \(5^{\circ}\mathrm{C}\). What is the conduction rate?

Solution. Apply \(P_{\text{cond}} = kA\,\Delta T/L\) directly.

P_cond = kA ΔT / L
\[ P_{\text{cond}} = \frac{(1.0)(1.5)(20 - 5)}{3.0 \times 10^{-3}} = 7.5 \times 10^{3}\,\mathrm{W} = 7.5\,\mathrm{kW} \]

An enormous rate — which is why single-pane glass is such a poor barrier. Double glazing inserts a thin air layer (\(k \approx 0.026\)); using the series rule \(P = A\,\Delta T/\sum(L/k)\), that air gap dominates the resistance and slashes the loss many-fold.

Review

Chapter Summary

Zeroth law & Kelvin scale

Equilibrium with a common body means equal temperature. Anchored at the triple point: \(T_{3} = 273.16\,\mathrm{K}\).

Temperature scales

\(T_{C} = T - 273.15^{\circ}\), \(T_{F} = \tfrac{9}{5}T_{C} + 32^{\circ}\). Scales meet at −40°.

Thermal expansion

\(\Delta L = L\alpha\,\Delta T\), \(\Delta V = V\beta\,\Delta T\), \(\beta = 3\alpha\). Holes expand too.

Heat & specific heat

\(Q = cm\,\Delta T\). Water: \(c = 4187\,\mathrm{J/kg\cdot K}\). Heat is energy in transit, not stored.

Phase changes

\(Q = Lm\). Water: \(L_{F} = 333\,\mathrm{kJ/kg}\), \(L_{V} = 2256\,\mathrm{kJ/kg}\).

Work by a gas

\(W = \int p\,dV\) — the area under the p–V curve. Path-dependent.

First law

\(\Delta E_{\text{int}} = Q - W\). Path-independent: it depends only on the states.

Special cases

Adiabatic \(\Delta E = -W\); const-V \(\Delta E = Q\); cycle \(Q = W\); free exp. \(\Delta E = 0\).

Heat transfer

Conduction \(P = kA\,\Delta T/L\); convection (buoyant fluid); radiation \(P = \sigma\varepsilon A T^{4}\).

Practice

Problems

Unless told otherwise, use \(c_{\text{water}} = 4187\,\mathrm{J/kg\cdot K}\), \(L_{F} = 333\,\mathrm{kJ/kg}\), \(L_{V} = 2256\,\mathrm{kJ/kg}\) for water, and \(\sigma = 5.67 \times 10^{-8}\,\mathrm{W/m^{2}\cdot K^{4}}\). Expansion problems hinge on \(\alpha\) (and \(\beta = 3\alpha\)); calorimetry on "heat gained = heat lost"; first-law problems on tracking the signs of \(Q\) and \(W\).

  1. The normal boiling point of nitrogen is 77.4 K. Express this on (a) the Celsius scale and (b) the Fahrenheit scale.
  2. At what single temperature do the Celsius and Fahrenheit scales give the same numerical reading? Verify your answer using \(T_{F} = \tfrac{9}{5}T_{C} + 32\).
  3. A steel railroad track is 12.0 m long when laid at 10 °C. By how much does it lengthen on a 40 °C day? (Steel: \(\alpha = 11 \times 10^{-6}/\mathrm{C^{\circ}}\).)
  4. A brass ring of diameter 10.00 cm at 20 °C is to slip over a steel rod of diameter 10.01 cm at 20 °C. To what common temperature must both be heated so the ring just fits the rod? (Brass: \(19 \times 10^{-6}\); steel: \(11 \times 10^{-6}\).)
  5. An aluminum cup holds 0.20 kg of water. How much heat raises both from 20 °C to 80 °C? (Aluminum cup mass 0.15 kg, \(c_{\text{Al}} = 900\,\mathrm{J/kg\cdot K}\).)
  6. How much heat must be removed from 1.50 kg of water at 20 °C to turn it entirely into ice at 0 °C?
  7. A 0.30 kg block of ice at 0 °C is dropped into 1.0 kg of water at 30 °C in an insulated container. Does all the ice melt? What is the final temperature?
  8. A gas expands from \(V_{i} = 1.0\,\mathrm{m^{3}}\) to \(V_{f} = 4.0\,\mathrm{m^{3}}\) at a constant pressure of \(2.0 \times 10^{5}\,\mathrm{Pa}\). How much work does the gas do?
  9. During a process a gas absorbs 400 J of heat and does 150 J of work on its surroundings. What is the change in its internal energy?
  10. A gas is compressed adiabatically, with 220 J of work done on it. By how much does its internal energy change, and in which direction?
  11. A gas is carried around the closed cycle shown by a clockwise loop on a p–V diagram, doing 80 J of net work. What is the net heat added over the cycle, and what is \(\Delta E_{\text{int}}\) for the cycle?
  12. A brick wall is 0.12 m thick (\(k = 0.71\,\mathrm{W/m\cdot K}\)), with area 15 m², inner face at 22 °C and outer face at 2 °C. Find the rate of heat loss by conduction.
  13. A composite wall has a 3.0 cm layer of wood (\(k = 0.11\)) backed by a 6.0 cm layer of fiberglass (\(k = 0.048\)), area 10 m², faces at 25 °C and −5 °C. Find the steady conduction rate using the series rule.
  14. A blackbody sphere (\(\varepsilon = 1\)) of radius 5.0 cm is held at 500 K in surroundings at 300 K. Find (a) the power it radiates and (b) the net power it exchanges with its surroundings.
Tip: three habits carry most of this chapter. For expansion, decide first whether you need linear (\(\alpha\)) or volume (\(\beta = 3\alpha\)) change. For calorimetry, write "heat lost = heat gained" and watch for phase changes hiding extra \(Lm\) terms. For the first law, fix your sign convention once — \(W\) is work done by the system — and then read every special case off \(\Delta E_{\text{int}} = Q - W\) by setting one term to zero.