10.3 Heat Transfer and Newton's Law of Cooling
Heat moves in three ways: conduction through matter ($H=KA\,\Delta T/L$), convection with moving fluid, and radiation across empty space ($P=e\sigma AT^4$). For small temperature differences all of them give Newton's law of cooling.
Stir a pan of boiling dal with a steel spoon and in a minute the handle is too hot to hold; a wooden spoon stays cool. Hold your hand above a candle flame and it burns, but beside the flame, at the same distance, it is only pleasantly warm. Step out of the shade into the sunshine and you feel the warmth instantly, though the energy has crossed $150$ million kilometres of empty space. Three quite different processes are at work: conduction along the spoon, convection as hot air rises above the flame, and radiation from the Sun.
This section looks at each in turn. For conduction we find a simple law for the rate of heat flow through a slab and combine slabs like resistors in series — the physics of insulating walls, copper-bottomed pans and double glazing. Convection we describe qualitatively, from sea breezes to the circulation in a pot. For radiation we meet the blackbody, the Stefan–Boltzmann law ($P\propto T^4$) and Wien's displacement law, which lets astronomers read a star's temperature from its colour. Finally, for small temperature differences, all three combine into Newton's law of cooling, the exponential approach to room temperature that the teaser of this chapter showed.
Conduction
In conduction, heat passes through a material without the material itself moving. In a hot region the atoms vibrate more vigorously and jostle their neighbours, passing energy along; in metals, free electrons carry energy much faster, which is why good electrical conductors are also good conductors of heat.
Take a slab of thickness $L$ and face area $A$, with one face held at $T_1$ and the other at a lower $T_2$. After a while the temperature at each point inside stops changing — the steady state. Heat then flows through at a constant rate $H$ (joules per second, i.e. watts), and experiment shows that $H$ is proportional to the area and the temperature difference and inversely proportional to the thickness.
In the steady state, the rate of heat flow by conduction through a uniform slab of area $A$ and thickness $L$, whose faces are at temperatures $T_1\gt T_2$, is
$$H=KA\,\frac{T_1-T_2}{L},$$from the hotter face to the colder. The constant $K$ is the thermal conductivity of the material; its unit is $\text{W m}^{-1}\text{K}^{-1}$. The quantity $\dfrac{T_1-T_2}{L}$ is the temperature gradient; in the steady state the temperature falls linearly through a uniform slab.
| Material | $K$ ($\text{W m}^{-1}\text{K}^{-1}$) | Material | $K$ ($\text{W m}^{-1}\text{K}^{-1}$) |
|---|---|---|---|
| copper | 385 | glass | 0.80 |
| aluminium | 205 | brick, water | 0.60 |
| steel | 50 | wood | 0.15 |
| ice | 2.2 | glass wool; air | 0.040; 0.024 |
Metals conduct hundreds of times better than glass or brick, and still air is one of the best insulators — which is why woollen clothes, feathers, glass wool and double-glazed windows all work by trapping air. A metal railing feels colder than wood at the same temperature because it conducts heat away from your hand much faster.
Thermal resistance. Rewrite the law as $H=\dfrac{T_1-T_2}{R}$ with $R=\dfrac{L}{KA}$. It reads exactly like Ohm's law, current $=$ voltage$/$resistance, with heat flow playing the current and temperature difference the voltage. For slabs placed one after another, the same $H$ must pass through each (otherwise heat would pile up at a junction and the state would not be steady), so their temperature drops add.
The thermal resistance of a slab is $R=\dfrac{L}{KA}$ (unit K/W). For slabs of the same area in series, in the steady state,
$$R=R_1+R_2+\cdots,\qquad H=\frac{T_\text{hot}-T_\text{cold}}{R},$$and the temperature drop across each slab is $HR_i$ — largest across the slab with the largest resistance. For two layers of equal thickness, the equivalent conductivity is $K_\text{eq}=\dfrac{2K_1K_2}{K_1+K_2}$.
Convection
Convection is the transfer of heat by the actual movement of a fluid (liquid or gas) from one place to another, carrying its energy with it. In natural convection the motion is driven by gravity: fluid heated from below expands, becomes less dense (Section 10.1) and rises, while cooler, denser fluid sinks to take its place. In forced convection the fluid is pushed by a fan or a pump.
Convection cannot happen in a solid, and natural convection needs gravity — on the International Space Station a candle flame is a dim sphere, not a tall flame, because the hot gases do not rise. Some examples:
- Heating a pot from below: the bottom water warms, rises, and is replaced by cooler water — a circulating current that warms the whole pot far faster than conduction alone could (water's $K$ is only $0.60\ \text{W m}^{-1}\text{K}^{-1}$).
- Sea and land breezes: by day the land heats faster than the sea (sand and rock have a much smaller specific heat capacity than water, Section 10.2), so air over the land rises and cool air flows in from the sea; at night the land cools faster and the breeze reverses.
- Trade winds and monsoons: on a planetary scale, hot air rising near the equator and cooler air flowing in from higher latitudes set up steady wind patterns; the seasonal heating of the Indian landmass drives the monsoon.
- Forced convection: the fan in a laptop, the water pump of a car engine, and the blood that the heart pumps from the warm body core to the skin.
This is why a room heater is placed low and an air conditioner high, and why the freezing compartment of an old refrigerator is at the top.
Radiation
Conduction and convection both need matter. Radiation does not: every body emits thermal radiation — electromagnetic waves, mostly infrared at everyday temperatures — because of the thermal motion of its charged particles. It travels at the speed of light, crosses a vacuum, and is absorbed (and turned back into internal motion) by whatever it falls on.
A blackbody is an ideal body that absorbs all the radiation falling on it, at every wavelength; it is also the best possible emitter at any given temperature. A real surface emits a fraction $e$ of what a blackbody of the same area and temperature would emit; $e$ (between $0$ and $1$) is its emissivity. Good absorbers are good emitters: a dull black surface has $e$ close to $1$, polished silver about $0.02$. A small hole in a hollow box, through which radiation enters and is absorbed after many internal reflections, is an excellent practical blackbody.
The power radiated by a body of surface area $A$, emissivity $e$ and absolute temperature $T$ is
$$P=e\sigma AT^4,$$with the Stefan–Boltzmann constant $\sigma=5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}$ ($e=1$ for a blackbody). A body at $T$ in surroundings at $T_s$ also absorbs radiation from them, so its net rate of loss is
$$P_\text{net}=e\sigma A\left(T^4-T_s^4\right).$$A blackbody at temperature $T$ emits over a continuous range of wavelengths, with an intensity peak at wavelength $\lambda_m$ given by
$$\lambda_mT=b,\qquad b=2.9\times10^{-3}\ \text{m K}.$$As a body gets hotter, its peak moves to shorter wavelengths: a heated iron rod glows dull red, then orange, then yellow-white.
The $T^4$ is dramatic: doubling the absolute temperature multiplies the radiated power by $16$. At room temperature radiation is comparable with convection; for a furnace, a lamp filament or a star it dominates everything else. A vacuum flask attacks all three routes at once: the vacuum between its double walls stops conduction and convection, and silvering the walls cuts radiation (low $e$ means poor emission and poor absorption).
Newton's law of cooling
A hot body in a cool room loses heat by all three routes at once. When its temperature $T$ is not far above the surroundings at $T_s$, each route gives a rate of loss roughly proportional to the excess $T-T_s$. For conduction and convection this is nearly so by experiment; for radiation it follows from the algebra
$$T^4-T_s^4=(T-T_s)(T+T_s)(T^2+T_s^2),$$which is $\approx4T_s^3(T-T_s)$ when $T\approx T_s$.
For a body whose temperature $T$ exceeds that of its surroundings $T_s$ by a small amount (typically a few tens of kelvin at most), the rate of loss of heat is proportional to the temperature excess:
$$-\frac{dQ}{dt}=k_1\left(T-T_s\right),$$where $k_1$ depends on the area and nature of the surface. If the body has mass $m$ and specific heat capacity $c$, then $dQ=mc\,dT$ and
$$\frac{dT}{dt}=-k\left(T-T_s\right),\qquad k=\frac{k_1}{mc}.$$The rate of cooling is greatest when the body is hottest and falls as it approaches $T_s$.
(a) Average-rate form. Over a short interval in which the temperature falls from $T_1$ to $T_2$ in time $t$,
$$\frac{T_1-T_2}{t}\approx k\left(\frac{T_1+T_2}{2}-T_s\right).$$(b) Exponential form. Exactly, for constant $k$ and $T_s$, the temperature excess decays exponentially (you will meet the integration that gives this in mathematics):
$$T-T_s=\left(T_0-T_s\right)e^{-kt},$$so a graph of $\ln(T-T_s)$ against $t$ is a straight line of slope $-k$, and the excess halves every $\dfrac{\ln2}{k}$.
1. Celsius in $T^4$. The Stefan–Boltzmann and Wien laws need kelvin. And $T^4-T_s^4$ is not $(T-T_s)^4$.
2. Thickness upside-down. $H\propto\dfrac{A}{L}$: a thicker wall lets less heat through.
3. Adding conductivities in series. In series, resistances $L/KA$ add, not conductivities.
4. Assuming the junction is at the average temperature. The larger temperature drop is across the layer of larger resistance (Example 2: $79\ ^\circ\text{C}$, not $50\ ^\circ\text{C}$).
5. Newton's law for large excesses. It is an approximation for small $T-T_s$; for a red-hot body radiation's $T^4$ dominates and the law fails.
6. Constant-rate cooling. Cooling slows as the body approaches room temperature: equal temperature drops take longer and longer (Example 6).
In JEE Main, NEET and JEE Advanced (as applications of heat conduction); not required by the CBSE syllabus.
Parallel slabs. When two slabs of the same thickness $L$ sit side by side between the same two temperatures (a wall with a window in it), each carries its own heat current and the currents add: $H=H_1+H_2$, so $\dfrac1R=\dfrac1{R_1}+\dfrac1{R_2}$, and for equal areas $K_\text{eq}=\dfrac{K_1+K_2}{2}$ — the rules of resistors in parallel.
Growth of ice. On a pond the water under the ice is at $0\ ^\circ\text{C}$ and the air above at $-\theta$. When the ice is $x$ thick, heat flows up through it at $H=\dfrac{KA\theta}{x}$ and freezes a further layer $dx$ in time $dt$: $\rho A\,dx\,L_f=\dfrac{KA\theta}{x}dt$. Adding up the layers (the area under the graph of $x$ against $x$), the time to thicken from $x_1$ to $x_2$ is
$$t=\frac{\rho L_f}{2K\theta}\left(x_2^2-x_1^2\right).$$Example. With $\rho=917\ \text{kg/m}^3$, $K=2.2\ \text{W m}^{-1}\text{K}^{-1}$, $L_f=3.33\times10^5\ \text{J/kg}$ and air at $-10\ ^\circ\text{C}$, growing from $5.0$ to $10\ \text{cm}$ takes $t=\dfrac{917\times3.33\times10^5}{2\times2.2\times10}$ $\times(0.010-0.0025)$ $=5.2\times10^4\ \text{s}\approx14\ \text{h}$ — three times as long as the first $5\ \text{cm}$, because the ice insulates the water below it.
Exercises
Use the conductivities in the table of this section, $\sigma=5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}$, $b=2.9\times10^{-3}\ \text{m K}$ and $L_f=3.33\times10^5\ \text{J/kg}$. All conduction problems are in the steady state, with lagged sides.
- 1 A copper rod $0.50\ \text{m}$ long, of cross-section $2.0\ \text{cm}^2$, has its ends at $100\ ^\circ\text{C}$ and $20\ ^\circ\text{C}$. Find the rate of heat flow.
- 2 A brick wall $4.0\ \text{m}\times2.5\ \text{m}$ and $0.20\ \text{m}$ thick has its inner surface at $22\ ^\circ\text{C}$ and outer surface at $2\ ^\circ\text{C}$. Find the rate of heat loss, and the energy lost in a day in kWh.
- 3 What thickness of glass wool has the same thermal resistance as the brick wall of part 2 (same area)?
- 4 Find the thermal resistance of a wooden door $2.0\ \text{m}\times0.90\ \text{m}$ and $2.0\ \text{cm}$ thick.
- 5 A wall of area $1.0\ \text{m}^2$ consists of $10\ \text{cm}$ of brick on the inside and $5.0\ \text{cm}$ of glass wool on the outside. The inner surface is at $25\ ^\circ\text{C}$ and the outer at $5\ ^\circ\text{C}$. Find the heat current and the temperature where the brick meets the glass wool.
- 6 Two rods of equal length and area, one copper and one steel, are joined in series. Find the conductivity of a single rod of the same total length that would conduct the same heat.
- 7 A metal tray and a wooden table have been in the same room all day. Why does the tray feel colder to the touch?
- 8 (Modelling) A polystyrene ice box ($K=0.033\ \text{W m}^{-1}\text{K}^{-1}$) has walls $2.0\ \text{cm}$ thick and a total wall area of $0.80\ \text{m}^2$. It contains ice at $0\ ^\circ\text{C}$ on a $30\ ^\circ\text{C}$ day. Find the rate of heat flow in and the mass of ice that melts each hour.
- 1 Explain why a breeze blows from the sea to the land during a sunny day and from the land to the sea at night.
- 2 Why is a room heater placed near the floor, and the cooling unit of an air conditioner near the ceiling?
- 3 The spectrum of a star peaks at $290\ \text{nm}$. Estimate its surface temperature and describe its colour compared with the Sun.
- 4 At what wavelength does the thermal radiation of a human body at $37\ ^\circ\text{C}$ peak? In which part of the spectrum is this?
- 5 By what factor does the power radiated by a blackbody increase when its temperature rises from $300\ \text{K}$ to $600\ \text{K}$?
- 6 Find the power radiated by a blackbody sphere of radius $5.0\ \text{cm}$ at $727\ ^\circ\text{C}$.
- 7 The tungsten filament of a lamp has surface area $0.50\ \text{cm}^2$ and emissivity $0.35$, and runs at $2500\ \text{K}$. Find the power it radiates.
- 8 Explain how each feature of a vacuum flask — double glass walls with a vacuum between them, silvered surfaces, an insulating stopper — reduces heat transfer.
- 1 A body cools from $60\ ^\circ\text{C}$ to $50\ ^\circ\text{C}$ in $10\ \text{min}$ in a room at $25\ ^\circ\text{C}$. Using the average-rate form, how long does it take to cool from $50\ ^\circ\text{C}$ to $40\ ^\circ\text{C}$?
- 2 For the same body, estimate its temperature $10\ \text{min}$ after it was at $50\ ^\circ\text{C}$.
- 3 Coffee at $85\ ^\circ\text{C}$ is left in a room at $25\ ^\circ\text{C}$, with $k=0.040\ \text{min}^{-1}$. Find its temperature after $15\ \text{min}$.
- 4 When does the coffee of part 3 reach $40\ ^\circ\text{C}$?
- 5 In a room at $25\ ^\circ\text{C}$, compare the rates of cooling of a body at $75\ ^\circ\text{C}$ and at $45\ ^\circ\text{C}$.
- 6 Show that for a small excess $\Delta T=T-T_s$, $T^4-T_s^4\approx4T_s^3\Delta T$. Taking $T_s=300\ \text{K}$ and $\Delta T=10\ \text{K}$, by what percentage is the approximation in error?
- 7 (Modelling) A forensic team finds a body in a room at $20\ ^\circ\text{C}$. Its temperature is $30.0\ ^\circ\text{C}$ at 11:00 pm and $28.0\ ^\circ\text{C}$ at midnight. Assuming Newton's law of cooling in exponential form and a normal body temperature of $37.0\ ^\circ\text{C}$, estimate the time of death.
- 1 A pan has a base $3.0\ \text{mm}$ thick and of area $0.040\ \text{m}^2$, and $1.5\ \text{kW}$ flows through it into boiling water. Find the temperature difference across the base if it is made of (a) steel, (b) copper. Why are good pans copper-bottomed?
- 2 Why are light-coloured clothes cooler in strong sunshine, and why are the outsides of cooking pots often left blackened?
- 3 A uniform lagged rod $2.0\ \text{m}$ long has one end at $100\ ^\circ\text{C}$ and the other at $0\ ^\circ\text{C}$. What is the steady temperature $0.50\ \text{m}$ from the hot end?
- 4 (Modelling) A hot-water tank with surface area $1.5\ \text{m}^2$ is wrapped in a jacket $5.0\ \text{cm}$ thick of material with $K=0.040\ \text{W m}^{-1}\text{K}^{-1}$. The water is at $60\ ^\circ\text{C}$ and the room at $20\ ^\circ\text{C}$. Estimate the heat lost per day, in kWh.
E1. Board · 3 marks State Newton's law of cooling and the condition under which it holds. Sketch the graph of temperature against time for a cooling body, and the graph of $\ln(T-T_s)$ against time.
E2. NEET The spectra of two stars, treated as blackbodies, peak at $350\ \text{nm}$ and $700\ \text{nm}$. The ratio of the power emitted per unit area by the first star to that by the second is (a) $2$ (b) $4$ (c) $8$ (d) $16$
E3. JEE Main (Numerical value) Rod A ($K=400\ \text{W m}^{-1}\text{K}^{-1}$, length $0.30\ \text{m}$) and rod B ($K=100\ \text{W m}^{-1}\text{K}^{-1}$, length $0.20\ \text{m}$), of equal cross-section, are joined end to end and lagged. The free end of A is at $100\ ^\circ\text{C}$ and that of B at $0\ ^\circ\text{C}$. Find the temperature of the junction in $^\circ\text{C}$, to one decimal place.
E4. JEE Advanced (One or more correct) Blackbody sphere A has radius $r$ and temperature $T$; blackbody sphere B has radius $2r$ and temperature $T/2$. Which statements are correct? (A) A radiates four times the total power of B. (B) The wavelength at which B's spectrum peaks is twice that of A. (C) A emits sixteen times as much power per unit area as B. (D) A and B emit equal total power because the larger area of B compensates for its lower temperature.
E5. Board · 1 mark JEE Advanced A liquid cools from $70\ ^\circ\text{C}$ to $60\ ^\circ\text{C}$ in $5\ \text{min}$ in surroundings at $30\ ^\circ\text{C}$. Using the average-rate form of Newton's law of cooling, the time it takes to cool from $60\ ^\circ\text{C}$ to $50\ ^\circ\text{C}$ is (a) $5\ \text{min}$ (b) $6\ \text{min}$ (c) $7\ \text{min}$ (d) $8\ \text{min}$
E6. Board · Case study A kitchen-ware company sells stainless-steel pans with a copper base and vacuum flasks with silvered double walls. (i) Why is copper used for the base? [1] (ii) Name the mode(s) of heat transfer each feature of the flask is designed to stop: the vacuum, and the silvering. [1] (iii) The copper base is $2.0\ \text{mm}$ thick and has area $0.030\ \text{m}^2$. Find the rate of heat flow through it when its two faces differ in temperature by only $0.20\ \text{K}$. [2]
Research question. How much cooler would a top-floor room be in an Indian summer if its roof were painted white?
Why it matters. Heatwaves are becoming longer and hotter across South Asia, the Middle East, Africa and southern Europe, and people in top-floor homes under bare concrete or metal roofs suffer most. Several Indian cities now run "cool roof" programmes as part of their heat action plans, and similar schemes exist in cities from Los Angeles to Athens. The physics is radiation (how much sunlight the roof absorbs) followed by conduction (how much of that heat reaches the room).
What to do. Model a flat concrete roof $0.15\ \text{m}$ thick ($K\approx1.4\ \text{W m}^{-1}\text{K}^{-1}$) over a room whose ceiling surface is at $30\ ^\circ\text{C}$. On a sunny afternoon the outer surface of a dark roof reaches about $65\ ^\circ\text{C}$, while a white-painted roof stays near $40\ ^\circ\text{C}$.
- Find the rate of heat conduction per square metre through each roof.
- For a $20\ \text{m}^2$ room, convert each to kilowatts and compare with the cooling power of a typical $1.5$-tonne air conditioner (about $5\ \text{kW}$).
- Explain, using absorption and emissivity, why white paint lowers the outer surface temperature. Why do the best cool-roof coatings also have a high emissivity in the infrared?
- Use Newton's law of cooling to argue how the room temperature would respond if the heat input were suddenly reduced in the evening.
- Find published measurements of indoor temperature reductions from cool-roof trials and compare them with your estimate.
Tools: Calculator or spreadsheet; an infrared (contactless) thermometer if available, to measure dark and light surfaces in the sun; reports from city heat action plans and building-energy research groups.
Hand in: Your calculations, a labelled diagram of the heat flows through each roof, and a one-page note to a residents' association recommending (or not) a white roof, with costs and benefits.
Going further. Combine a cool coating with a layer of insulation under the roof: treat the two as slabs in series and find which change gives the bigger reduction per rupee spent.
🧪 Self-check quiz
Q1. The SI unit of thermal conductivity is
Q2. Energy from the Sun reaches the Earth by
Q3. The absolute temperature of a blackbody is doubled. The power it radiates becomes
Q4. According to Newton's law of cooling, the rate of cooling of a body is proportional to
Q5. Two slabs of equal area are in series. The first has thermal resistance $R$, the second $2R$. In the steady state, the temperature drop across the first slab is