Heat

HEAT TRANSFER (return to CONTENTS)


MECHANISMS (return to start of page)

There are three main mechanisms of heat transfer:

Conduction

Observation 1:

If a metal spoon is dipped into a cup of hot water, the end of the spoon soon feels warm. Heat is transferred along the spoon by conduction.

When a point on an object is heated, the molecules there vibrate more strongly. Conduction is the transfer of these vibrations within the material.

In a metal there are many millions of ‘free electrons’ which make the process of conduction fairly rapid.
Insulators (= poor conductors) such as wood or plastic do not have free electrons and conduction is very slow.

Suppose that a piece of steel and a piece of wood are both at room temperature (about 200C) and you put a hand (at about 370C) on each. Heat flows from your hand more rapidly into the steel, a good heat conductor, which therefore feels colder than the wood

Observation 2:

This indicates that water is a poor conductor of heat. Gases are also poor conductors.

Convection

This is a much quicker means of heat transfer in fluids (i.e. gases and liquids) than conduction.

Observation:

As before, the test tube is held with tongs over the flame as shown, and the open end should not point towards anyone. In convection the actual material moves, carrying heat with it - unlike in conduction.

Natural convection

i) Sea breeze

ii) Land breeze
Forced convention - Newton’s law of cooling

If an object is cooling under forced convention, i.e. subject to a steady draught, Newton’s law of cooling applies:

Radiation

Observation:

When the heater is turned on, the cork stuck to the blackened aluminium foil falls off within seconds.

Surfaces which look black do so because they absorb visible light, whereas shiny surfaces reflect visible light.

The heater glows red because it emits visible red light. But it also emits invisible infrared radiation ('heat radiation'). This is also absorbed by the black surface, so the foil quickly becomes warm, melting the wax, and the cork drops off.

Note: Black surfaces absorb radiation better than shiny surfaces and they also emit radiation better than shiny surfaces - hence, the blackened surfaces at the back of a fridge.

Conduction, convection and radiation compared

Conduction and convection both require material mediums. Conduction is the transfer of atomic/molecular vibrations. Convection is the motion of the fluid itself, carrying heat with it. Thus, neither of these can occur in a vacuum.

Radiation can travel though a vacuum, i.e. it does not require a material medium. If this were not so, we would not exist because most life on earth depends upon energy from the sun, and this comes in the form of visible and invisible radiation that travels through about 150 million km (93 million miles) of (mostly) empty space to reach the earth.


BLACK BODY RADIATION (return to start of page)

A black surface is a good absorber of electromagnetic ('em') radiation. A ‘black body’ in physics is a perfect absorber - it absorbs all the radiation falling on it, reflecting none. The opening in an empty can of pop looks black from most directions, and is quite a good approximation to a black body. The light entering undergoes multiple reflections before finding the hole again, and loses some intensity at each reflection, so an emerging ray is much weaker than an incident ray:

This would be an even better black body if the inside surface were coated with matt (=dull) black.

A black body radiator (also called a cavity radiator) is one that emits radiation which is characteristic of the temperature of the body. The above can could be made into a black body radiator by surrounding it with a heating coil. It could be kept at a chosen temperature and the radiation emerging from the hole could be analysed.


Wien’s displacement law


The constant = 2.9 * 10-3 mK ('metres kelvin')  [Note: kelvins, K = oC+ 273]

For the sun, the wavelength at which the most intense radiation is emitted is in the visible spectrum. For visible light, the wavelength ranges from 4.0*10-7m to 7.0*10-7m. Using the average of these, 5.5*10-7m, the above equation implies that the surface temperature of the sun is about 5300K.

Stefan’s law


Planck’s theory (1900)

In trying to explain the shape of the above graphs, Max Planck suggested that light, and all other electromagnetic radiation, of frequency f, is emitted in ‘packets’ of energy called ‘quanta’, of size E given by:


Planck’s idea was later used successfully by Einstein to explain a completely different phenomenon.


THERMAL CONDUCTIVITY (return to start of page)

Definition of thermal conductivity, k

Suppose Q joules of heat pass through a thin slab of material in time t seconds:


Unit of k = W/(m2 0C m-1) = W m-1  0C-1 or W m-1 K-1

From equation (2), we can define thermal conductivity as:

Heat flow along a well-lagged metal bar

The following represents an insulated bar from which no heat is lost from the sides:

The top diagram represents a well-lagged metal bar with its ends maintained at different temperatures. In practice, the hot end may be in a steam chest, while the cold end may be in melting ice. After sufficient time the temperatures shown by the thermometers stay constant. When this occurs we say that ‘steady state’ has been reached. When the steady state has been reached, a graph as shown above can be plotted.

The graph represents the steady state variation of temperature along the bar. If the bar were poorly insulated, heat would escape from the sides, and the steady state graph would be a curve. In the above graph, the temperature gradient equals the slope, and is given by:


Some prefer to express this as:

Example

  1. Find the rate of flow of heat through a plaster ceiling which measures 4m x 3m x 10mm, if the inside temperature is 150C and the outside temperature is 50C.
  2. The same ceiling is now insulated with a 50mm layer of fibre-glass. Find (i) the temperature at the interface between the ceiling and the fibre-glass, and (ii) the rate of flow of heat through the ceiling.
(for plaster, kp = 0.6 Wm-1K-1; for fibre-glass, kf = 0.04 Wm-1K-1)

In this case, the area through which heat flows is very large (several m2) compared to the thickness it flows through, so there will be very little heat loss from the edges relative to the amount flowing through the ceiling, so equation (1) above applies (or versions (3) or (4)):

(x is used in place of  l, to represent thickness, in the following)



Now, the rate of flow of heat through both layers must be the same. If, say, there were more heat per second flowing though the plaster than through the fibre-glass, then heat would be building up at the boundary, and this does not occur.

This is the rate of flow though the fibre-glass and the plaster ceiling. Thus, the fibre-glass makes a significant difference to the rate of heat loss.


U-VALUES (return to start of page)

Example

Estimate the rate of flow of heat through a window if the inside temperature is 200C and the outside temperature is 50C. The area of the window is 2m2, and it is made of 6mm thick glass.
(for glass, k = 0.8 Wm-1K-1)

This value is unrealistically high. In practice the rate of heat loss is much less than this. In the previous problem we saw that the rate of heat loss fell considerably when the insulating fibre-glass was in place. A window also has some (invisible) insulation, in the form of a layer of air near each surface. Air is a poor conductor, (k=0.02 Wm-1K-1), and so each layer acts like the fibre-glass in producing a large reduction in heat loss.

The calculation above assumes that the temperature of the window surfaces is that of the air, and implies a ‘temperature profile’ across the air-glass-air to be like:

A more realistic temperature profile is:
This indicates a larger temperature gradient across the air layers than the glass.

To take into account practical considerations, such as the insulating effect of invisible layers of air, heating engineers measure actual rates of heat loss through parts of buildings and quote the ‘U-value’ for a particular window, a particular door etc.

As an equation we would express this as:


Example

An outside wall of a room has an area of 10.0 m2. It is made of two thicknesses of brickwork separated by an insulating foam. Find the rate at which heat escapes through the wall if the inside temperature is 200C, the outside temperature is 50C, and the wall has a U-value of 0.50 Wm-2K-1.



COMPARISON OF HEAT AND CHARGE TRANSFER (return to start of page)

If a charge Q flows through R in time t, then:

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A Level Physics - Copyright © A C Haynes 1999 & 2004