Heat always flows from a region of higher temperature to a region of lower
temperature. It flows by **conduction**,
**convection**, and **radiation**. Often we are interested in regulating the rate at which
the thermal energy is transferred. We may want to keep an object at a
temperature different from that of its surrounding for a long time by slowing
down heat flow. Or we may want an object to cool down rapidly by increasing the
rate at which thermal energy is transferred. In developing methods to do this
effectively, we always have to consider the importance of the three different
ways by which heat flows.

If we surround an object at temperature T_{2} with a layer of
material, to insulate it from its surrounding at temperature T_{1}, then
the **thermal conductivity** of the surrounding
material determines how fast heat can flow through it.

- Let ΔT = (T
_{2}- T_{1}) be the difference in temperature between side 2 and side 1 of a layer of material of area A. - Let Δx be the thickness of that layer.
- Let ΔQ/Δt be the amount of heat that flows from side 1
to side 2 through the layer of material per unit time.

(A negative ΔQ/Δt indicates that heat flows from side 2 to side 1.)

The **thermal conductivity k** is defined through the equation
ΔQ/Δt = -kA ΔT/Δx.

Thermal conductivities:

(kcal/sec)/(^{o}C m)

Aluminum | 4.9 * 10^{-2} |
---|---|

Copper | 9.2 * 10^{-2} |

Steel | 1.1 * 10^{-2} |

Air | 5.7 * 10^{-6} |

Ice | 4 * 10^{-4} |

Wood | 2 * 10^{-5} |

Glass | 2 * 10^{-4} |

Asbestos | 2 * 10^{-5} |

The unit kcal (kilocalorie) is a unit of energy.

1 kcal = 4186 J.

This equation is called the **law of heat conduction**. ΔQ/Δt
is the rate at which heat flows across the area A, in Joules per second
or Watts. ΔT/Δx is the change in the temperature over the distance
Δx in degrees Kelvin or Celsius per meter. It is the **temperature
gradient**. The thermal conductivity k is a property of the
material.

To minimize heat flow through a layer of material by
conduction, choose the right material, make the layer as thick as possible, and
make the surface area as small as possible.

**Clothing** intended to reduce heat flow
should use low-thermal conductivity materials. Clothes should be relatively
thick and trap air, since air is a poor thermal conductor. The materials
should not contain metals, since metals are good thermal conductors.

Calculate the rate of heat flow (in J/s = Watts) by conduction through a
glass window 2.0 m multiplied by 3 m in area and 4 mm thick, if the inside
temperature is 15^{o} C and the outside temperature is -5^{o} C.

Solution:

Let the inside be side 2 and the outside be side 1.

ΔQ/Δt = -kA*(T_{2} - T_{1})/Δx

= (-2*10^{-4})[(kcal/sec)/(^{o}C m)]*(4186 J/kcal)*(2 m * 3
m)*(20 ^{o}C)/(4*10^{-3} m)

= -25116 J/s = -25 kilowatt.

The minus sign indicates that the heat flows from the inside to the outside.

This is an enormous rate of heat flow. The next problem shows how by trapping a thin layer of air between two layer of glass we can reduce heat loss by conduction significantly.

A Thermopane window of area 6 m^{2} is constructed of two layers of
glass, each 4 mm thick, separated by an air space of 5 mm. If the inside is at
15 ^{o}C and the outside is at -5 ^{o}C, what is the rate of
heat loss through the window? The thermal conductivity of the glass is 0.84 W/(m^{o}C)
and the thermal conductivity of the air is 0.0234 W/(m^{o}C).

Solution:

When a steady state is reached, then the same amount of heat crosses any
cross sectional area per second. Let the temperature of the inner
glass-air boundary be T_{1} and the temperature of the outer
glass-air boundary be T_{2}. Then for the inner piece of glass
we have

-ΔQ/Δt = (0.84 W/(m^{o}C))*6 m^{2}*(15 ^{o}C - T_{1})/0.004
m.

For the air layer we have

-ΔQ/Δt = (0.0234 W/(m^{o}C))*6 m^{2}*(T_{1
}- T_{2})/0.005
m.

For the outer piece of glass we have

-ΔQ/Δt = (0.84 W/(m^{o}C))*6 m^{2}*(T_{2
}+ 5 ^{o}C)/0.004
m.

The first and third equation yield (15 ^{o}C - T_{1}) = (T_{2
}+ 5^{o}C), T_{2 }= 10^{o}C - T_{1}.

Inserting this expression for T_{2} into the second equation we have

-ΔQ/Δt = (0.0234 W/(m^{o}C)*6 m^{2}*(2 T_{1} - 10
^{o}C)/0.005 m.

Combining this equation with the first equation then yields

(0.0234 W/(m^{o}C))*(2 T_{1} - 10
^{o}C)/(0.005 m) =
(0.84 W/(m^{o}C))*(15^{o }C - T_{1})/(0.004 m).

0.0223*(2 T_{1} + 10
^{o}C) = 15 ^{o}C - T_{1}.

1.045 T_{1 }= 14.8
^{o}C. T_{1 }= 14.1 ^{o}C.

The first equation now yields

-ΔQ/Δt = (0.84 W/(m^{o}C)*6 m^{2}*(0.85
^{o}C)/0.004 m = 1.1 kW.

A bar of gold
is in thermal contact with a bar of silver of the same length and area. One
end of the compound bar is maintained at 80 ^{o}C and the opposite
end is at 30 ^{o}C. When the heat flow reaches steady state, find
the temperature at the junction,

Solution:

When a steady state is reached, then the same amount of heat
crosses any cross sectional area perpendicular to the bar per second.
(Otherwise the energy would increase in certain regions, the temperature
would increase there, the temperature gradient would change, and we would
not have a steady state.) If the temperature at the junction is T, then in
the gold we have

ΔQ/Δt = (314 W/(m^{o}C))*A*(80 ^{o}C -
T)/(L/2)

and in the silver we have

ΔQ/Δt = (427 W/(m^{o}C))*A*(T
- 30 ^{o}C)/(L/2).

We can therefore write

(314 W/(m^{o}C))*A*(80
^{o}C - T)/(L/2) = (427 W/(m^{o}C))*A*(T - 30^{o}C)/(L/2).

(314 W/(m^{o}C))*(80 ^{o}C - T) = (427W/(m^{o}C))*(T
- 30 ^{o}C).

(80 ^{o}C - T) = 1.36*(T - 30 ^{o}C).

120.8 ^{o}C = 2.36 T.

T = 51.2 ^{o}C.

Heat naturally flows from a region of higher to a region of
lower temperature. In fluids a hotter region has lower density than a colder
region. Near the surface of the earth, where the gravitational acceleration
points downward, buoyancy causes the hotter fluid to rise, setting up convection
currents. The rate of heat flow depends on the heat capacity and mobility of
the fluid, i.e. how quickly heat flows into or out of the fluid and how
well the fluid circulates because of buoyancy.

Buoyancy is not always moving a fluid effectively. It fails when the hotter
fluid is above the colder fluid, when fluids experience large drag forces, or
when the geometry of the container prevents fluid flow. One can help the fluid
move and therefore enhance heat transfer by stirring the fluid. In air, wind
increases heat transfer (wind chill) by forcing convection.

**Clothing** can reduce convective heat transfer
by preventing fluids from circulating.

- Fluffy clothing traps air and prevents convection (geometry drag).
- Thick clothing allows the surface temperature of the clothing to drop to that of your surrounding air and thus prevents external convection.
- A wind breaker minimizes forced convection.

The particles that make up an object can have ordered energy and
disordered energy. The temperature is a measure of this internal,
disordered energy. The **absolute temperature **of any substance is
proportional to the average kinetic energy associated with the random motion of
the substance.

The velocity of particles with thermal energy is changing almost
all the time. The particles are **accelerating**. The atoms and molecules
are themselves complicated arrangements of charged particles. Accelerating
charged particles produce electromagnetic radiation. The power radiated is
proportional to the square of the acceleration. Higher rates of velocity change
result in higher frequency (shorter wavelength) radiation.

Experimental challenges:

- Radiation carries away energy and therefore cools the object.
- Some of the radiation coming from the object may be reflected or transmitted radiation.

Ideal solution:

Use a **blackbody.** A blackbody is a body
that absorbs all the radiation that falls onto it. It does not reflect any
radiation. It reaches thermal equilibrium with its surroundings, and in thermal
equilibrium emits exactly as much radiation it absorbs. It has emissivity
= 1. **Emissivity**** **measures the fraction of
radiative energy that is absorbed by the body.

Experimental realization:

Use the inside of a large box (oven) held at a constant temperature T. A small hole is cut in one side. Any radiation that enters through the hole bounces around inside and has little chance of ever getting out again. Eventually it gets absorbed. Radiation coming out the hole is as good a representation of the radiation from a perfect emitter.

What is observed?

The observed intensity of the radiation emitted as a function of wavelength
can be described by the **Planck Radiation Law.**

The Planck Radiation Law gives the
intensity of radiation emitted by a blackbody
as a function of wavelength for a fixed temperature. The Planck law gives a
distribution, which peaks at some wavelength. The peak shifts to shorter
wavelengths for higher temperatures, and the area under the curve grows rapidly
with increasing temperature. The diagram below shows the intensity distribution
predicted by the Plank law in J/(m^{2}s) for blackbodies at various
temperature.

The **Wien law** and the **Stefan-Boltzmann Law** can be
derived from the Planck Radiation Law.

The Wien Law gives the
wavelength of the peak of the radiation distribution,

λ_{max }= 3*10^{6}/T.

Here λ is measured
in units of nanometer = 10^{-9 }m and T is in Kelvin.

The Wien law can be the basis of a
non-contact measurement of the temperature of a hot object.

One measures the wavelength distribution of the radiation emitted by
the object and from the peak infers the temperature.

Link: Disappearing-filament pyrometer

The Wien law explains the **shift of the peak** to shorter
wavelengths as the temperature increases.

The Stefan-Boltzmann Law
gives the total energy being emitted at all wavelengths by the body.

Radiated power = emissivity * σ * T^{4} * Area

Here σ is the **Stefan-Boltzmann constant**,

= 5.67*10^{-8 }W/(m^{2}K^{4}) and the
temperature is measured in Kelvin.

The Stefan-Boltzmann
law explains the **growth in the height of the curve** as the
temperature increases. This growth is very abrupt, since it
varies as the fourth power of the temperature.

Link: PhET Simulation: Blackbody Spectrum

- The surface temperature of the sun is 5800
^{o}C = 6073 K. The wavelength of the peak of the distribution is 494 nanometer. This wavelength lies in the yellow region of the visible spectrum. - In an incandescent light bulb a filament is heated to approximately 2500
^{o}C = 2773K. This is the maximum temperature that a tungsten filament can stand without evaporating quickly. Compared to the sun, such a filament emits a greater fraction of its radiation in the infrared region of the electromagnetic spectrum. The wavelength of the peak of the distribution is 1082 nanometer. This wavelength lies in the infrared region of the spectrum. - Sunlight and light from an incandescent bulb contain all the colors of the visible spectrum. But the intensity distribution over the different colors is different. Sunlight appears brilliant white while a light bulb looks yellowish.

The **emissivity** of an object is the ratio
of the radiative power emitted by that object at temperature T to the radiative
power emitted by a blackbody of the same shape and temperature T. It is also
equal to the fraction of the incoming blackbody radiation at that temperature
that is not reflected but absorbed by the object. The emissivity of objects is
a function of the temperature T and can be quite different for visible and
infrared radiation.

The emissivity of an object for radiation emitted by high temperature sources (visible light) is easily determined by just looking at the object. A black surface has a high-temperature emissivity near 1, while a white or shiny surface has a high-temperature emissivity near 0. Light-colored or reflective objects have low emissivity. They do absorb a smaller percentage of the incoming visible radiation than do dark objects, and also emit radiation less readily.

One cannot "see" the emissivity of an object for radiation emitted by low temperature sources (infrared light). Most materials have a low-temperature emissivity near 1, but conducting (metallic) surfaces can have a low-temperature emissivity near 0. If you wrap a hot object in reflective aluminum foil, then the foil reflects most of the radiation emitted by the object back onto the object. Heat loss by radiation is thus slowed down.

To reduce radiative heat transfer, use low-emissivity surfaces and allow exterior surfaces to reach the ambient temperature.

A burning log is a black object with a surface area of 0.25 m^{2} and
a temperature of 800 ^{o}C. How much power does it emit as thermal
radiation?

Solution:

Assume the emissivity is close to 1.

Radiated power = emissivity * σ * T^{4} *
Area

= (5.67*10^{-8} J/(s m^{2} K^{4}))*(1073)^{4}*(0.25
m^{2}) = 18790 J/s.

Note: The temperature is measured in Kelvin.

Compare the rate R_{1} at which cup of water at 50^{o}C
radiates energy with he rate R_{2} at which cup of water at 100^{o}C
radiates energy/

Solution:

R_{2}/R_{1} = (273 + 100)^{4}/(273 + 50)^{4}
= 1.78

(Remember that the radiated power is proportional to the 4th power of the
absolute temperature.)