Conduction, convection and radiation
On a cold January morning in school in Cork, you sit beside a radiator. The metal feels hot to the touch, the air above it feels warm, and your face feels warm even when you are a metre away. One radiator, three different ways for energy to reach you. This page explains each one and shows how to put numbers on them. At the end we will come back to the radiator.
Conduction
Conduction is the transfer of thermal energy through a material without the material itself moving.
Use the particle model. In the hot end of a metal spoon the particles vibrate with large amplitude. They knock against their neighbours, which then vibrate more strongly, and so on down the spoon. Energy is passed along, but the particles stay where they are.
Metals are far better conductors than wood or plastic. A metal also has free (delocalised) electrons that move through the whole solid and carry energy quickly. Wood has no free electrons, so it relies on vibrations alone and conducts poorly.
How fast does energy flow?
Take a bar of length $L$ and cross-sectional area $A$, with its ends at different temperatures. The rate of energy flow depends on four things:
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a bigger area $A$ gives a bigger flow (more particles take part)
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a larger temperature difference $\Delta T$ gives a bigger flow
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a longer bar $L$ gives a smaller flow (the energy has further to go)
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the material matters, and this is measured by its thermal conductivity $k$
$Q/t$ is in watts ($\text{W}$), $k$ in $\text{W m}^{-1}\text{K}^{-1}$, $A$ in $\text{m}^2$, $\Delta T$ in $\text{K}$ (or $^\circ\text{C}$, since it is a difference) and $L$ in $\text{m}$. A big $k$ means a good conductor. Copper has $k \approx 400$, while a good insulator such as mineral wool has $k \approx 0.04$.
A metal spoon left in a mug of hot tea in a kitchen in Galway becomes warm at the end that is not in the tea. a) Name the process by which thermal energy travels along…
A metal spoon left in a mug of hot tea in a kitchen in Galway becomes warm at the end that is not in the tea.
- a) Name the process by which thermal energy travels along the spoon.
- b) Using the particle model, describe how energy is passed from the hot end to the cool end of the spoon.
- c) Explain why a metal spoon warms up faster than a wooden spoon of the same size.
A copper rod of length $0.50\ \text{m}$ and cross-sectional area $2.0\times10^{-4}\ \text{m}^2$ has one end in boiling water at $100\,^\circ\text{C}$ and the other end…
A copper rod of length $0.50\ \text{m}$ and cross-sectional area $2.0\times10^{-4}\ \text{m}^2$ has one end in boiling water at $100\,^\circ\text{C}$ and the other end in an ice bath at $0\,^\circ\text{C}$. The thermal conductivity of copper is $k = 400\ \text{W m}^{-1}\text{K}^{-1}$. Assume no energy is lost from the sides of the rod and that the rate of energy flow is given by $\dfrac{Q}{t}=\dfrac{kA\Delta T}{L}$.
- a) Calculate the rate of energy flow along the rod.
- b) Calculate the energy transferred in $5.0$ minutes.
- c) The copper rod is replaced by a steel rod of identical dimensions with $k = 50\ \text{W m}^{-1}\text{K}^{-1}$. Calculate the new rate of energy flow.
A hot-water tank in a house in Cork is cylindrical. Heat is lost to the surroundings mainly by conduction through the insulating jacket. The jacket has area…
A hot-water tank in a house in Cork is cylindrical. Heat is lost to the surroundings mainly by conduction through the insulating jacket. The jacket has area $2.4\ \text{m}^2$, thickness $0.080\ \text{m}$ and thermal conductivity $0.040\ \text{W m}^{-1}\text{K}^{-1}$. The water inside is at $60\,^\circ\text{C}$ and the air outside is at $18\,^\circ\text{C}$.
- a) Calculate the rate of heat loss through the jacket.
- b) Calculate the energy lost in $24$ hours, in $\text{kWh}$.
- c) If electricity costs €0.34 per $\text{kWh}$, calculate the daily cost of replacing this lost energy.
Convection
Convection is the transfer of thermal energy by the movement of a fluid (a liquid or a gas). It cannot happen in a solid, because the particles in a solid are fixed in position and cannot flow.
Follow what happens over a radiator:
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The air next to the radiator is heated. Its particles move faster and spread out.
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The same mass of air now takes up a larger volume, so its density decreases.
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The less dense warm air is pushed upwards by the denser cool air around it (it experiences an upthrust).
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Cool air flows in to take its place, is heated in turn, and a convection current is set up.
The same idea explains why the hot water tank stores its hottest water at the top: the heated water is less dense, so it rises above the cooler water below.
Sea breezes
Land and sea receive the same sunshine, but the sea heats up much more slowly. Water has a specific heat capacity of about $4{,}180\ \text{J kg}^{-1}\text{K}^{-1}$, whereas dry land is about $800\ \text{J kg}^{-1}\text{K}^{-1}$. The same energy raises the temperature of the land much more. Also, the sea mixes by convection, so the energy is shared through a deep layer of water.
By day the land is warmer, so the air above it rises and cooler air flows in from the sea. This is the onshore breeze you feel at Salthill. At night the land cools faster than the sea. Now the air over the sea is warmer and rises, and the breeze reverses.
A storage heater in a classroom in Limerick warms the air next to it. The warm air rises and cooler air moves in to replace it. a) Name this method of heat transfer. b)…
A storage heater in a classroom in Limerick warms the air next to it. The warm air rises and cooler air moves in to replace it.
- a) Name this method of heat transfer.
- b) Explain, in terms of density and the motion of particles, why the warm air rises.
- c) Explain why convection cannot occur in a solid.
A sea breeze often blows onshore on a summer afternoon at Salthill, Galway. Land heats up faster than the sea during the day. a) Explain why the land heats up faster…
A sea breeze often blows onshore on a summer afternoon at Salthill, Galway. Land heats up faster than the sea during the day.
- a) Explain why the land heats up faster than the sea, referring to specific heat capacity.
- b) Using the idea of convection currents, explain why the breeze blows from the sea to the land during the day.
- c) Explain why the direction of the breeze often reverses at night.
Radiation
Thermal radiation is energy carried by electromagnetic waves, mainly infrared. It needs no medium, so it crosses the vacuum between the Sun and the Earth. Every object above $0\ \text{K}$ emits it, and it travels at $3.0\times10^{8}\ \text{m s}^{-1}$.
What happens to radiation arriving at a surface depends on the surface:
| Surface | Absorbs | Reflects | Emits (when hotter than surroundings) |
|---|---|---|---|
| Dark, matt | a lot | little | a lot |
| Light, shiny | little | a lot | little |
A good absorber is also a good emitter. That is why a white house stays cooler in summer sunshine, and why a hot water bottle cover or a kettle is often shiny, to lose less energy by radiation.
The Stefan–Boltzmann law
The power radiated by a perfect emitter (a black body) depends strongly on its absolute temperature:
$$P=\sigma A T^{4}$$where $A$ is the surface area, $T$ is the temperature in kelvin, and $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}$. Doubling $T$ makes the power $2^4 = 16$ times bigger.
An object also absorbs radiation from its surroundings at temperature $T_s$. The net power lost is
$$P_{net}=\sigma A\left(T^{4}-T_s^{4}\right)$$On a sunny day, a student standing on Dollymount Strand in Dublin feels the warmth of the Sun on her face. a) Name the method by which energy travels from the Sun to the…
On a sunny day, a student standing on Dollymount Strand in Dublin feels the warmth of the Sun on her face.
- a) Name the method by which energy travels from the Sun to the Earth.
- b) Explain why this method can transfer energy through the vacuum of space, whereas conduction and convection cannot.
- c) State why a white house reflects more of this energy than a dark-coloured house does.
A black-painted metal cube of side $0.10\ \text{m}$ is at $400\ \text{K}$ in a laboratory at $300\ \text{K}$. It radiates as a black body, and the Stefan–Boltzmann law…
A black-painted metal cube of side $0.10\ \text{m}$ is at $400\ \text{K}$ in a laboratory at $300\ \text{K}$. It radiates as a black body, and the Stefan–Boltzmann law gives the power emitted as $P=\sigma A T^4$, where $\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}$.
- a) Calculate the total surface area of the cube.
- b) Calculate the power radiated by the cube.
- c) Calculate the power the cube absorbs from the surroundings at $300\ \text{K}$, and hence the net rate of energy loss by radiation.
- d) The cube has mass $8.0\ \text{kg}$ and specific heat capacity $900\ \text{J kg}^{-1}\text{K}^{-1}$. Estimate the initial rate of fall of its temperature, and state why this is only an estimate.
A student investigates which of two identical cans of water cools faster: one painted matt black and one polished silver. Each can contains $0.25\ \text{kg}$ of water…
A student investigates which of two identical cans of water cools faster: one painted matt black and one polished silver. Each can contains $0.25\ \text{kg}$ of water initially at $90\,^\circ\text{C}$ in a room at $20\,^\circ\text{C}$. After $10$ minutes the black can is at $62\,^\circ\text{C}$ and the silver can is at $71\,^\circ\text{C}$. The specific heat capacity of water is $4,180\ \text{J kg}^{-1}\text{K}^{-1}$. Ignore the heat capacity of the cans.
- a) Calculate the energy lost by each can in $10$ minutes.
- b) Calculate the average power lost by each can.
- c) Explain the difference in the results in terms of emission of radiation.
- d) The experiment is repeated with each can filled with water and a lid placed on top. Predict and explain the effect on the results, referring to convection and evaporation.
Layers in series
A real wall or window has several layers. Energy must pass through every layer one after another. In the same way that resistors add in a series circuit, thermal resistances add.
For one layer of thickness $L$, area $A$ and conductivity $k$:
$$R=\frac{L}{kA}$$The unit is $\text{K W}^{-1}$. Rewriting the conduction formula gives $\dfrac{Q}{t}=\dfrac{\Delta T}{R}$. For layers in series, the same rate of flow passes through each layer and
$$R_{total}=R_1+R_2+R_3+\dots \qquad \frac{Q}{t}=\frac{\Delta T}{R_{total}}$$A layer of poorly conducting material (like air or foam) has a large $R$ and dominates the total. That is why double glazing works: the trapped air layer provides most of the resistance. It also helps that a thin trapped layer of air cannot set up large convection currents.
A double-glazed window consists of two glass panes, each $4.0\ \text{mm}$ thick with $k_g = 1.0\ \text{W m}^{-1}\text{K}^{-1}$, separated by a $12\ \text{mm}$ layer of…
A double-glazed window consists of two glass panes, each $4.0\ \text{mm}$ thick with $k_g = 1.0\ \text{W m}^{-1}\text{K}^{-1}$, separated by a $12\ \text{mm}$ layer of air with $k_a = 0.026\ \text{W m}^{-1}\text{K}^{-1}$. The window has area $1.5\ \text{m}^2$. The inside surface is at $20\,^\circ\text{C}$ and the outside surface is at $2\,^\circ\text{C}$. Treat the heat flow as steady conduction through the three layers in series, using thermal resistance $R=\dfrac{L}{kA}$.
- a) Calculate the thermal resistance of each layer and the total thermal resistance.
- b) Calculate the rate of heat flow through the window.
- c) Calculate the rate of heat flow through a single $4.0\ \text{mm}$ pane of the same area under the same temperature difference, and comment on the result.
- d) Explain why the air gap is a much better insulator than glass, referring to the particle model and to convection.
Back to the start: the radiator in your classroom. The hot metal warms your hand by conduction. The air above it is heated and rises as a convection current. Your face is warmed by infrared radiation, which travels straight through the air.
Summary
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Conduction: energy passes from particle to particle in a solid. Metals are the best conductors because of free electrons. $\dfrac{Q}{t}=\dfrac{kA\Delta T}{L}$.
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Convection: energy is carried by a moving fluid. Heated fluid expands, becomes less dense and rises. It cannot happen in solids.
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Radiation: infrared and other electromagnetic waves. It needs no medium, and dark matt surfaces absorb and emit best. $P=\sigma A T^4$, with $T$ in kelvin.
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Net radiated power: $P_{net}=\sigma A(T^4-T_s^4)$.
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Layers in series: $R=\dfrac{L}{kA}$, $R_{total}=R_1+R_2+\dots$ and $\dfrac{Q}{t}=\dfrac{\Delta T}{R_{total}}$.
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Energy lost: $E = (Q/t)\times t$. To convert to kWh, use power in kW and time in hours.