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Unformatted text preview: Lesson 1.5 Thermal Properties of Matter. Lesson Objectives: At the end of this lesson students will be able to (i) know and apply thermal properties of matter to explain thermal expansion and transfer of heat energy. (ii) use the thermal properties of matter to solve problems . 1. Thermal Expansion. Heat supplied to a material increases its internal energy, and causes the molecules to vibrate with larger amplitudes. As a result, each molecule now takes up a little more space than before and causes the material to expand. As a result of the expansion, its length, area and volume will change. We will consider each of these changes separately. (i) Change in length : When an object in the form of a rod is heated, its increase in length ∆ L is proportional to its original length L o . This can be written as ∆ L = α L o ∆ T …1 where α is the constant of proportionality and is called the coefficient of linear expansion and is equal to the change in length per unit length per Kelvin change in temperature. o L L T α ∆ = ∆ …2 It has the unit, per Kelvin (K1 ) If L 1 is the length of a rod at t 1 o C and L 2 is its length at t 2 o C, then ) ( 1 2 1 1 2 t t L L L = α …3 This can be rearranged to obtain an equation for L 2 as: L 2 = L 1 [1 + α (t 2 – t 1 )] …4 (ii) Change in area : Similar expressions can be written for change in area with temperature as follows: ∆ A = β A ∆ T where β is the constant of proportionality and is called the coefficient of area expansion. It is a measure of the increase in area per unit area per Kelvin rise in temperature. ) ( 1 2 1 1 2 t t A A A T A A = ∆ ∆ = β …5 A 2 = A 1 [1 + β (t 2 – t 1 )] ….6 Since an area is made up of two lengths, an approximate relation between α and β can be written as β = 2 α …7 (iii) Change in volume : Similar expressions can be written for the change in volume as follows: ∆ V = γ V ∆ T …8 ) ( 1 2 1 1 2 t t V V V T V V = ∆ ∆ = γ …9 V 2 = V 1 [1 + γ (t 2 – t 1 ) …10 Since an increase in volume is caused by increase in length in three directions, an approximate relation connecting α and γ can be written as: γ = 3 α …11 The following table gives the values of α and γ of some common materials. Material α K1 γ K1 Aluminum 24 × 106 75 × 106 Brass 19 × 106 56 × 106 Copper 17 × 106 50 × 106 Iron or Steel 11 × 106 35 × 106 Lead 29 × 106 87 × 106 Ordinary glass 9 × 106 18 × 106 Gasoline 950 × 106 Mercury 180 × 106 Water 210 × 106 Air 3400 × 106 Since an increase in temperature causes an object to expand, the same mass occupies a greater volume, and so the density of the material decreases. An approximate relation between density of a substance at 0 o C (d o )and its density at t o C (d t ) can be written as d t = d o (1  γ t) …12 Example 1 : A steel rule has a length of 30 cm at 20 o C. What is its length at 100 o C? ( α steel = 11 × 106 K1 ) Solution: L 1 = 30 cm = 0.3 m, t 1 = 20 o C, t...
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This note was uploaded on 05/01/2011 for the course PHY 2049 taught by Professor George during the Spring '11 term at Edison State College.
 Spring '11
 George
 Physics, Energy, Heat

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