BMAT Physics Notes: Thermal Physics

Advice & Insight From BMAT Specialists

Transfer of Heat

There are three ways that heat can travel from one place to another: conduction, convection and radiation.
CONDUCTION:  This is the method that heat travels through a solid. Liquids (apart from mercury) and gases are very poor conductors of heat.
In a solid, the atoms are vibrating in fixed  positions. If one end of a metal bar is heated then the atoms vibrate with increased frequency and amplitude. This causes the neighbouring atoms to vibrate faster and so on. In this way heat energy is transferred through the material. Note that the atoms stay in their positions.
All metals are good conductors of heat. Non-metals are poor conductors (insulators) of heat.
CONVECTION:  If a liquid or gas is heated the atoms move further apart and so the liquid or gas expands. This causes its density to decrease and so is less than the surrounding (cooler) liquid/gas. This causes the warm liquid or gas to rise. Cooler denser liquid/gas then moves across to take its place. This in turn is heated, expands, becomes less dense and rises.  In this way a current, called a convection current is set up in the liquid/gas and eventually all the liquid/gas is heated.
Note that, unlike conduction in a solid, the atoms do move and exchange positions with other atoms.
(THERMAL) RADIATION:  Both conduction and convection require a medium for heat to travel through.
The Earth gets most of its heat from the Sun. Since there is no medium between the Earth and the Sun there must be another method that heat can be transferred. This method is called (thermal) radiation.
Any hot body will emit em radiation called infra-red waves. It is by this process that heat energy can travel through a vacuum (or through a medium).
Dark objects are good absorbers/emitters of infra-red waves.
Light white objects are poor absorbers/emitters of infra-red waves.
Shiny objects are good reflectors of infra-red waves.
[NOTE: do NOT confuse thermal radiation with nuclear radiation].

Specific Heat Capacity

 Def: The specific heat capacity (c) is the amount of heat required to raise the temperature of 1kg of a substance by 1°C (or 1 K)
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Equation for heat gained/lost

Heat gained/lost  =  mass x s.h.c. x temperature change
                        or,  Q = m  c  θ
where  Q = heat lost/gained,  m = mass (in kg),  c = specific heat capacity,  θ = temperature change.

Water has a specific heat capacity ≈ 4200 J kg⁻¹ K⁻¹
This means 4200 J is required to raise the temperature of 1 kg of water by 1° C

Example 1

​How much heat is required to raise the temperature of 2kg of water from 20°C to boiling point?

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Example 2

How much heat is lost when 0.5kg of water cools from 800 C to 20° C ?​

Example 3

A block of metal of density 6000 kg m⁻³ has dimensions of 0.4m x 0.5m x 0.8m and has a s.h.c. of 400 J kg⁻¹ K⁻¹
Calculate the heat required to raise its temperature from  -20° C to + 20° C ?

Example 4
Calculate the specific heat capacity of a lump of metal of mass 0.1 kg if 5000 J of heat causes its temperature to rise from  -5 °C to 120°C ?

Example 5

An electrical storage heater contains a heating element, thermal blocks and insulation as shown in the diagram.

The heating element heats the thermal blocks during the night.
(a) Between midnight and 6.00 a.m,  8.64 x 107 J of energy are supplied to the heating element. 
​     (i) Calculate the power rating of the heating element.
    (ii) The total mass of the thermal blocks in the heater is 144 kg and the specific heat capacity of the thermal blocks is 2625 J kg⁻¹ K⁻¹. Calculate the maximum possible rise in the temperature of the thermal blocks between midnight and 6.00 a.m.
    (iii) Explain why the actual temperature rise of the blocks is less than the value calculated above.  
    (iv) Why is there insulation between the thermal blocks and the outer casing of the heater ?

Example Answers

​Example 1:   H = m c θ  =  2 x 4200 x 80  =  6.72 x 10 J

​Example 2:   H = m c θ  =  0.5 x 4200 x 60  =  1.26 x 10 J

Example 3:   Density = mass/volume   mass = density x vol = 6000 x 0.4 x 0.5 x 0.8 = 960 kg
                           H = mcθ = 960 x 400 x 40 = 1.54 x 10J

Example 4:   c  =  H/mθ  =  5000/(0.1×125) =  400 J kg⁻¹ K⁻¹

Example 5:  Power =  Energy/time  =  8.64×10/(6×3600) =  4000 W
                          θ =  H/mc  = 8.64×10/(144×2625) = 229 °C
                           Heat loss to surroundings
                           Safety, casing may get too hot

BMAT Physics Notes: Thermal Physics

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