PHY 103 Lecture Notes 2020-2021 PDF

lOMoARcPSD|33130136 PHY 103 Lecture NOTE 2020 2021 General Physics 1 (Afe Babalola University) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by Donald Ngorka ([email protected]) ...

lOMoARcPSD|33130136 PHY 103 Lecture NOTE 2020 2021 General Physics 1 (Afe Babalola University) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 0 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 PROPERTIES OF MATTER (PHY103) Module 1: Elasticity 3 Hooke's law Young, shear and bulk moduli Module 2: Hydrostatics 9 Density and specific gravity Pressure Buoyance and Archimedes' Principles Surface tension; adhesion, cohesion and capillarity. Module 3: Temperature and Heat 19 Temperature Thermal Expansion Heat capacity and Phase Change Module 4: Heat Transfer 26 Conduction Convention Radiation Module 5: laws of Thermodynamics 32 Module 6: Kinetic Molecular Theory of Gasses 38 Gas Laws Kinetic Theory of Gases 1 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 INTRODUCTION Matter is anything that has weight and occupy space. Matter is normally classified as being in one of 3 states/phases: Solids, liquid or gas (or a combination of any of these) depending on the temperature and pressure. From everyday experience, we know that a solid has a definite volume and shape, liquid has a definite volume but no definite shape and finally, we know that an unconfined gas has neither a definite volume nor definite shape. These characteristics help picture the states of matter, but they are somewhat artificial. For example, asphalt and plastics are normally considered solids but turn liquids under special condition. Solids, liquids and gases are made up of tiny particles called molecules which are in random motion. The tip of a needle has many millions of molecules of the metal. When a solid, for example metal absorbs energy, the molecules vibrate/oscillate continuously about their mean positions. Let’s think of molecules in solids as miniature balls connected by springs that represent forces and bonds between them and the balls can vibrate through great distances as they steadily increase in amplitude (expansion). The increase in amplitude of vibration in a solid is small making expansion in solids to be generally small. When the energy absorbed by a solid material is great enough to make a molecules oscillate through a distance that permanently stretches the connecting “spring”, then a liquid form of the solid material is obtained. In liquids the molecules are not confined to one position as they move about inside the liquid. Their freedom notwithstanding, the liquid has a fixed volume. The molecules of gas have kinetic energy greater than that of liquid. It therefore moves in different directions throughout its volume in a container. The forces and bonds between molecules of gas are so weak that the molecules can move independently of each other. Gas molecules are obtained after an increase in the temperature of liquid further weakens the already stretched bonds between molecules in liquid. What will happen when you take away energy from a material in gaseous phase? 2 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 MODULE ONE ELASTICITY Elasticity is the ability of a substance to regain its original shape and size after the distorting external force has been removed e.g. suspension of a load from the end of a coil of spring, the spring stretches certain length, when the load is removed the spring returns to its former length. 1.1 Elastic Material Material that regains its original shape and size after the distorting force has been removed. The shape and size of a material can be distorted by either compressing or stretching it. And as soon as the compressing or stretching force is removed, an elastic material regains its original size and shape. 1.1.1 Elastic Limit: Elastic Limit is the border line of force beyond which the stretched wire does not return to its original length when the stretching force is removed. i.e. beyond the elastic limit the material losses its elasticity. 1.1.2 Yield point: Yield point is the point beyond the elastic limit in which the elastic material has yielded all its elasticity permanently and become plastic. 1.1.3 Breaking point: Breaking point is the maximum extension reached where the material may finally snap or break entirely. 3 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 Figure 1.1: The graph of extension against load up to breaking point where, P= Proportionality limit, E = Elastic limit, Y = Yield point, B = Breaking point, OE = Elastic region, EB = Plastic region 1.2 Hooke’s Law Hooke’s law is the law that governs the relationship between the stretching force and extension produced in the elastic material. When a load is attached to one end of a fixed string at a certain stage, Hooke’s law is obeyed. A point is later reached at which if more load is added, such that the load added is not proportional to the extension or compression of the material, then there will a breakdown, this point is called the BREAKING POINT. Therefore, Hooke’s Law states that, provided the elastic limit of an elastic material is not exceeded, the extension (e) of the material is directly proportional to the load or applied force (F). Mathematically, 𝑭𝑭 ∝ 𝒆𝒆 𝑭𝑭 = 𝑲𝑲𝒆𝒆 𝟏𝟏. 𝟏𝟏 Where 𝑲𝑲 is a constant of proportionality called the elastic constant 4 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 1.3 Elastic Properties of Solids 1.3.1 Stress and Strain Stress is defined as force per unit area on which it acts. Mathematically, 𝑭𝑭 𝑺𝑺𝑺𝑺𝑺𝑺𝒆𝒆𝑺𝑺𝑺𝑺 = 𝟏𝟏. 𝟐𝟐 𝑨𝑨 Strain is a measure of the degree of deformation. i.e. extension (𝒆𝒆) divided by the original length 𝒍𝒍𝟎𝟎 Mathematically, 𝒆𝒆 𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺 = 𝟏𝟏. 𝟑𝟑 𝒍𝒍𝟎𝟎 It is found that, for sufficiently small stress, strain is proportional to stress. The constant of proportionality depends on the material being deformed and on the nature of deformation. We call this proportionality constant the ELASTIC MODULUS (E.M). The E.M. is the ratio of the stress to the resulting strain. Mathematically, 𝑺𝑺𝑺𝑺𝑺𝑺𝒆𝒆𝑺𝑺𝑺𝑺 𝑬𝑬. 𝑴𝑴 = 𝟏𝟏. 𝟒𝟒 𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺 1.4 Young’s Modulus: Elasticity in Length Consider a long bar of cross- sectional area 𝑨𝑨 and initial length 𝒍𝒍𝑺𝑺 that is clamped at one end. When an external force is applied perpendicular to the cross- sectional area, internal forces in the bar resist distortion (stretching), but the bar attains an equilibrium in which its final length 𝒍𝒍𝒇𝒇 is greater than 𝒍𝒍𝑺𝑺 and the external force is exactly balanced by internal forces. In such a situation, the bar is said to be stressed. 5 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 Therefore, we define the tensile stress or compressive stress as the ratio of the magnitude of the external force 𝑭𝑭 to the cross-sectional area. The tensile strain or compressive strain in this case is defined as the ratio of the extension by ∆𝒍𝒍 to the original length, 𝒍𝒍𝑺𝑺. Tensile and compressive strains are defined in the same way but ∆𝒍𝒍 are in opposite directions. Young modulus has the same values for both tensile and compressive stresses. So, we define Young modulus (Y) by a combination of these two ratios: 𝑺𝑺𝑺𝑺𝑺𝑺𝒆𝒆𝑺𝑺𝑺𝑺 𝑭𝑭/𝑨𝑨 𝒀𝒀 = = 𝟏𝟏. 𝟓𝟓 𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺 ∆𝒍𝒍/𝒍𝒍𝑺𝑺 Young’s modulus is typically used to characterize a rod or wire stressed under either tension or compression. Its S.I. unit N/m2. 1.5 Shear Modulus: Elasticity of Shape Another type of deformation occurs when an object is subjected to a force tangential to one of its faces while the opposite face is held fixed by another force. If the object is originally a rectangular block, a shear stress results in a shape whose cross- section is a parallelogram. Therefore, shear stress is defined as the ratio of the tangential force to the area 𝑨𝑨 of the face being sheared. ∆𝑥𝑥 The shear strain is defined as 𝒉𝒉 Where, ∆𝒙𝒙 is the horizontal distance that the sheared face moves and 𝒉𝒉 is the height of the object. Therefore, shear modulus is 𝑺𝑺𝒉𝒉𝒆𝒆𝑺𝑺𝑺𝑺 𝑺𝑺𝑺𝑺𝑺𝑺𝒆𝒆𝑺𝑺𝑺𝑺 𝑭𝑭/𝑨𝑨 𝑺𝑺 = = 𝟏𝟏. 𝟔𝟔 𝑺𝑺𝒉𝒉𝒆𝒆𝑺𝑺𝑺𝑺 𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺 ∆𝒙𝒙/𝒉𝒉 6 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 1.6 Bulk Modulus: Volume Elasticity Bulk modulus characterizes the response of a substance to uniform squeezing or to a reduction in pressure when the object is placed in a partial vacuum. An object subjected to this type of deformation might change its volume by ∆𝑽𝑽 but no change in shape. The volume stress is stress experienced by a body when forces acts on all sides of the body. Now, if the pressure in an object changes by an amount ∆𝑷𝑷 then the object will experience a change in volume by ∆𝑽𝑽. The volume strain is the ratio of the change in volume by ∆𝑽𝑽 to the initial volume 𝑽𝑽𝑺𝑺 ∆𝑽𝑽 i.e. 𝑽𝑽 Thus, we can characterize a volume (bulk) compression in terms of a bulk modulus, which is defined as 𝑩𝑩𝑩𝑩𝒍𝒍𝑩𝑩 𝑺𝑺𝑺𝑺𝑺𝑺𝒆𝒆𝑺𝑺𝑺𝑺 ∆𝑷𝑷 𝑩𝑩 = = − 𝟏𝟏. 𝟕𝟕 𝑩𝑩𝑩𝑩𝒍𝒍𝑩𝑩 𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺 ∆𝑽𝑽/𝑽𝑽𝑺𝑺 The reciprocal of bulk modulus (k) is called compressibility which is defined as 𝟏𝟏 𝑩𝑩 = 1.8 𝑩𝑩 Compressibility is the fractional decrease in volume per unit increase in pressure. 7 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 1.7 Solved Problems 1. The tendon in a man’s leg is 10 𝑐𝑐𝑐𝑐 long and 0.45 𝑐𝑐𝑐𝑐 in diameter. How much will it be stretched by force of 5 𝑁𝑁 if the young’s modulus for the tendon is 1.6 𝑋𝑋 108 𝑁𝑁/𝑐𝑐2 ? Solution: 𝑙𝑙 = 10 𝑐𝑐𝑐𝑐 = 0.1 𝑐𝑐 𝐹𝐹 = 5 𝑁𝑁 𝑌𝑌 = 1.6 𝑋𝑋 108 𝑁𝑁/𝑐𝑐2 𝐷𝐷𝐷𝐷𝐷𝐷𝑐𝑐𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 0.45 𝑐𝑐𝑐𝑐 = 0.0045 𝑐𝑐 𝐷𝐷𝐷𝐷𝐷𝐷𝑐𝑐𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 0.0045 𝐷𝐷 = = = 0.00225 𝑐𝑐 2 2 22 𝐴𝐴 = 𝜋𝜋𝐷𝐷 2 = 𝑋𝑋 (0.00225)2 = 1.59 𝑋𝑋 10−4 𝑐𝑐 7 𝑻𝑻𝒆𝒆𝑺𝑺𝑺𝑺𝑺𝑺𝒍𝒍𝒆𝒆 𝑺𝑺𝑺𝑺𝑺𝑺𝒆𝒆𝑺𝑺𝑺𝑺 𝑭𝑭/𝑨𝑨 𝑭𝑭 𝒍𝒍𝑺𝑺 𝒀𝒀 = = = 𝑻𝑻𝒆𝒆𝑺𝑺𝑺𝑺𝑺𝑺𝒍𝒍𝒆𝒆 𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺 ∆𝒍𝒍/𝒍𝒍𝑺𝑺 𝑨𝑨 ∆𝒍𝒍 𝐹𝐹 𝑙𝑙𝑖𝑖 5 𝑋𝑋 0.1 ∆𝑙𝑙 = = = 𝟏𝟏. 𝟗𝟗𝟕𝟕 𝑿𝑿 𝟏𝟏𝟎𝟎−𝟒𝟒 𝒎𝒎 𝑌𝑌 𝐴𝐴 1.59 𝑋𝑋 10−4 𝑋𝑋1.6 𝑋𝑋 108 2. A specimen of oil having an initial volume of 500 𝑐𝑐𝑐𝑐3 is subjected to a pressure of 106 𝑁𝑁/𝑐𝑐2 and the volume decreases by 0.15 𝑐𝑐𝑐𝑐3. What is the bulk modulus for the oil? Solution: 𝑃𝑃 = 106 𝑁𝑁/𝑐𝑐2 𝑉𝑉𝑖𝑖 = 500 𝑐𝑐𝑐𝑐3 ∆𝑉𝑉 = 0.15 𝑐𝑐𝑐𝑐3. 𝑻𝑻𝒆𝒆𝑺𝑺𝑺𝑺𝑺𝑺𝒍𝒍𝒆𝒆 𝑺𝑺𝑺𝑺𝑺𝑺𝒆𝒆𝑺𝑺𝑺𝑺 𝑭𝑭/𝑨𝑨 𝑷𝑷 106 𝑩𝑩 = = = = 𝑻𝑻𝒆𝒆𝑺𝑺𝑺𝑺𝑺𝑺𝒍𝒍𝒆𝒆 𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺 ∆𝑽𝑽/𝑽𝑽𝑺𝑺 ∆𝑽𝑽/𝑽𝑽𝑺𝑺 (0.15/500) 106 = = 𝟑𝟑. 𝟑𝟑 𝑿𝑿 𝟏𝟏𝟎𝟎𝟗𝟗 𝑵𝑵/𝒎𝒎𝟐𝟐 3 𝑋𝑋 10−4 1.8 Supplementary Problems 8 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 1. A metal wire 75.0 𝑐𝑐𝑐𝑐 long and 0.130 𝑐𝑐𝑐𝑐in diameter stretches 0.0350 𝑐𝑐𝑐𝑐 when a load of 8.00 𝐾𝐾𝐾𝐾 is hung on its end. Find the stress, the strain and the Young’s modulus for the material of the wire 2. A solid cylindrical steel column is 4.0 𝑐𝑐𝑐𝑐 long and 9.0 𝑐𝑐𝑐𝑐 in diameter. What will be its decrease in length when carrying a load of 80000 𝐾𝐾𝐾𝐾? 𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝐾𝐾′ 𝑠𝑠 𝑐𝑐𝑌𝑌𝑚𝑚𝑌𝑌𝑙𝑙𝑌𝑌𝑠𝑠 (Y) = 1.9 𝑋𝑋 1011 𝑃𝑃𝐷𝐷 3. A bulk modulus of water is 2.1 𝐺𝐺𝑃𝑃𝐷𝐷. Compute the volume contraction of 100 𝑐𝑐𝑙𝑙 of water subjected to a pressure of 1.5 𝑀𝑀𝑃𝑃𝐷𝐷. 4. By how much will a wrought iron bar 0.006 𝑐𝑐2 in cross section area and 2 𝑐𝑐 long shorten under a compressive load of 2500 𝑁𝑁, if the Young’s modulus of wrought iron is 1.83 𝑋𝑋 1011 N/m2? 5. When a 400 𝐾𝐾 mass is hung at the end of a vertical spring, the spring stretches 35 𝑐𝑐𝑐𝑐. What is the spring constant of the spring, and how much further will it stretch if an additional 400 𝐾𝐾 mass is hung from it? 6. A hydraulic press contains 250 L of oil. Find the decrease in the volume of oil when it is subjected to a pressure of 1.6 x 107 Pa. The bulk modulus of oil is 5.0 x 109 Pa and its compressibility is 20 x 10-6 atm-1. 7. The brass plate of an out-door sculpture experiences shear forces in an earthquake. The plate is 0.80 m square and 0.50 cm thick. What is the force exerted on each of its edges if the resulting displacement x is 0.16 mm? (shear modulus = 3.5 x 1010 Pa). 9 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 MODULE TWO HYDROSTATICS Hydrostatics is the study of fluid mechanics (liquid and gas) when this fluid is in motion it is called hydrodynamics. Liquids and gases which alter their shapes in response to an applied force or to match the shape of its container are called fluids. 2.1 Density and Relative Density Density is the mass (m) of a unit volume (v) of a substance 𝒎𝒎 𝝆𝝆 = 𝟐𝟐. 𝟏𝟏 𝒗𝒗 The unit of density (𝝆𝝆) is 𝑲𝑲𝑲𝑲/𝒎𝒎𝟑𝟑. The density of water is given as 1000𝑘𝑘𝐾𝐾⁄𝑐𝑐3. Relative density (also known as specific gravity) of a substance is defined as the ratio of the density of that substance (𝝆𝝆) to the density of water (𝝆𝝆𝒘𝒘 ) i.e 𝒅𝒅𝒆𝒆𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝒅𝒅 𝒐𝒐𝒇𝒇 𝑺𝑺 𝑺𝑺𝑩𝑩𝒔𝒔𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝒔𝒔𝒆𝒆 𝝆𝝆 Relative density (R.D) = = 2.2 𝒅𝒅𝒆𝒆𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝒅𝒅 𝒐𝒐𝒇𝒇 𝒘𝒘𝑺𝑺𝑺𝑺𝒆𝒆𝑺𝑺 𝝆𝝆𝒘𝒘 It should be noted that R.D of water is 1.0 and a R,D is a ratio of density therefore it is always dimensionless. 2.1.1 Pressure A Fluid do not sustain shearing stress or tensile stress, thus the only stress that can be exerted on an object submerged in a fluid is the one that tends to compress the object, in other words the force exerted by a fluid in an object is always perpendicular to the surface of the object. Consider a container which is filled to height h with a liquid of density (𝝆𝝆) in Figure 2.1. In response to gravity, the liquid presses in the base of the container, the downward push is distributed evenly over the entire base. The pressure (P) on the surface of area A is defined as the normal force (F) on the surface divided by the area (A) of the surface. 10 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 𝑭𝑭 𝒎𝒎𝑲𝑲 𝝆𝝆𝑽𝑽𝑲𝑲 𝑷𝑷 = = = = 𝝆𝝆𝑲𝑲𝒉𝒉 𝟐𝟐. 𝟑𝟑 𝑨𝑨 𝑨𝑨 𝑨𝑨 Liquid (density) h P Figure 2.1: Pressure in a liquid The unit of pressure is 𝑁𝑁⁄𝑐𝑐2 or Pascal (Pa) Where, 1𝑁𝑁⁄𝑐𝑐2 = 1pa Standard atmospheric pressure is i.e. 1 atmosphere (atm) = 1.013 𝑋𝑋 105 𝑃𝑃𝐷𝐷 and 133.32 𝑃𝑃𝐷𝐷 = 1 𝑐𝑐𝑐𝑐 𝑌𝑌𝑜𝑜 𝑐𝑐𝐷𝐷𝐷𝐷𝑐𝑐𝑌𝑌𝐷𝐷𝑚𝑚 (𝑐𝑐𝑐𝑐𝑚𝑚𝐾𝐾) = 1 𝐷𝐷𝑌𝑌𝐷𝐷𝐷𝐷. Pressure depends only on the density and the height of the liquid and not on the shape of the container or the total volume of liquid. 2.2 Solved Problems 1. The mattress of a water bed is 2m long by 2m wide and 30cm deep. Find the weight of the water in the mattress. Solution: 𝜌𝜌𝑤𝑤 = 1000 𝑘𝑘𝐾𝐾⁄𝑐𝑐3 Length = 2m Wideness = 2m Depth = 30 cm = 0.3 m 𝒎𝒎 Recall that, 𝝆𝝆 = , 𝒗𝒗 11 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 Where, volume = Length X Width X Depth = 2 X 2 X 0.3 = 1.20 𝑐𝑐3 m = 𝝆𝝆𝒗𝒗 = (1000𝑘𝑘𝐾𝐾⁄𝑐𝑐3 )(1.20𝑐𝑐3 ) = 1.20 × 103 kg Also, W = mg = (1.20 × 103 kg)(9.80𝑐𝑐⁄𝑠𝑠 2 ) = 1.18 × 𝟏𝟏𝟎𝟎𝟒𝟒 N 2. Find the pressure due to the fluid at a depth of 76 cm in still (a) water (𝜌𝜌𝑤𝑤 = 1.00 𝐾𝐾/𝑐𝑐𝑐𝑐3 ) and (b) mercury (𝜌𝜌 = 13.6 𝐾𝐾/𝑐𝑐𝑐𝑐3 ) Solution: 𝜌𝜌 = 13.6 𝐾𝐾/𝑐𝑐𝑐𝑐3 = 13600 𝑘𝑘𝐾𝐾/𝑐𝑐3 𝜌𝜌𝑤𝑤 = 1.00 𝐾𝐾/𝑐𝑐𝑐𝑐3 = 1000 𝑘𝑘𝐾𝐾/𝑐𝑐3 ℎ = 76 𝑐𝑐𝑐𝑐 = 0.76 𝑐𝑐 𝐾𝐾 = 9.8 𝑐𝑐/𝑠𝑠 2 (a) 𝑷𝑷 = 𝝆𝝆𝒘𝒘 𝑲𝑲𝒉𝒉 = 1000 𝑋𝑋 9.8 𝑋𝑋 0.76 = 𝟕𝟕𝟒𝟒𝟓𝟓𝟎𝟎 𝑵𝑵/𝒎𝒎𝟐𝟐 (b) 𝑷𝑷 = 𝝆𝝆𝑲𝑲𝒉𝒉 = 13600 𝑋𝑋 9.8 𝑋𝑋 0.76 = 𝟏𝟏. 𝟎𝟎𝟏𝟏 𝑿𝑿 𝟏𝟏𝟎𝟎𝟓𝟓 𝑵𝑵/𝒎𝒎𝟐𝟐 2.3 Supplementary Problems 1. Atmospheric pressure is about 1.0 X 105 Pa. How large a force does the still air in a room exert on the inside of a window pane that is 40 cm X 80 cm? 2. You have just purchased a chain claimed to be pure gold. The chain weighs 60 g and it displaces 4.0 cm3 of water when fully immersed. Is it pure gold? (r.d. of gold = 19.3) 3. How high would water rise in the pipes of a building if the water pressure gauge shows the pressure at the ground floor to be 270 KPa. 4. A man’s brain is approximately 0.33 m above his heart. If the density of human blood is 1.05 X 103 Kg/m3, determine the pressure required to circulate blood between the heart and the brain. 12 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 5. Find the mass and weight of the air at 20oC in a living room with 4.0 m x 5.0 m floor and a ceiling 3.0 m high, and the mass and weight of equal volume of water. 2.4 Variation of Pressure with Depth Pressure increase with depth likewise, atmospheric pressure decrease with increasing altitude (height), it is for this reason that aircraft flying at high altitude must have pressurized cabins. A liquid which is confined in a container and an additional pressure is exerted on the surface of the liquid. An examples is the hydraulic press with the aid of piston the additional pressure is transmitted to every point of the liquid. In other words, the pressure exerted on the contained. Liquid is transmitted unchanged to every portion of the liquid. This is known as Pascal principle; this principle is utilized in the operation of hydraulic press. Pascal’s Principle state that when the pressure of any part of a confined fluid (liquid or gas) is changed, the pressure on every other part of the fluid is also changed by the same amount. Figure 2.2: Hydraulic Press The hydraulic press which consists essentially of a liquid bearing container with two necks fitted with movable pistons of cross section area 𝑨𝑨𝟏𝟏 𝐷𝐷𝑌𝑌𝑚𝑚 𝑨𝑨𝟐𝟐 (where𝑨𝑨𝟐𝟐 > 𝑨𝑨𝟏𝟏 ). Upon applying a force 𝑭𝑭𝟏𝟏 on the small piston, pressure 𝑷𝑷𝟏𝟏 = 𝑭𝑭𝟏𝟏 /𝑨𝑨𝟏𝟏 is generated which is transmitted to the large piston. 13 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 𝑭𝑭 The latter thus feels an upward force (𝑭𝑭𝟐𝟐 ) = 𝑷𝑷𝟏𝟏 𝑿𝑿 𝑨𝑨𝟏𝟏 or ( 𝟏𝟏 )𝑨𝑨𝟐𝟐 i.e. 𝑨𝑨𝟏𝟏 𝑭𝑭𝟐𝟐 𝑭𝑭𝟏𝟏 𝑷𝑷𝟏𝟏 = 𝑷𝑷𝟐𝟐 , = 𝑨𝑨𝟐𝟐 𝑨𝑨𝟏𝟏 𝑭𝑭𝟏𝟏 𝑨𝑨𝟐𝟐 𝑭𝑭𝟐𝟐 = 𝟐𝟐. 𝟒𝟒 𝑨𝑨𝟏𝟏 The hydraulic press is therefore a force multiplier with the mechanical advantage of 𝑨𝑨𝟐𝟐 𝑨𝑨 𝟏𝟏 2.5 Solved Problems In a car lift used in a service station compressed air exerts a force on a small piston that has circular cross section and radius of 5.00 cm. This pressure is transmitted by a liquid to the piston that has a radius of 15.0 cm. What force must the compressed air exert to lift a car weighing 13300 N what air pressure produces this force. Solution: 𝑺𝑺𝟏𝟏 = 5.0 𝑐𝑐𝑐𝑐 = 0.05 𝑐𝑐 𝑺𝑺𝟐𝟐 = 15.0 𝑐𝑐𝑐𝑐 = 0.15 𝑐𝑐 𝑭𝑭𝟐𝟐 = 13300 N 𝑭𝑭𝟏𝟏 = ? 𝑭𝑭 1.33× 104 𝑭𝑭𝟏𝟏 = 𝟐𝟐 𝑨𝑨𝟏𝟏 = (𝜋𝜋(0.05 )2 ) = 𝟏𝟏. 𝟒𝟒𝟒𝟒 × 𝟏𝟏𝟎𝟎𝟑𝟑 𝐍𝐍 𝑨𝑨 𝟐𝟐 𝜋𝜋(0.15 )2 (b) The air pressure that produce the force is 𝑭𝑭𝟏𝟏 1.48 × 103 𝑷𝑷𝟏𝟏 = = = 𝟏𝟏. 𝟒𝟒𝟒𝟒 × 𝟏𝟏𝟎𝟎𝟓𝟓 𝑷𝑷𝑺𝑺 𝑨𝑨𝟏𝟏 𝜋𝜋(5.00 × 10−2 )2 14 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 Supplementary Problems 1. The area of a piston of a force pump is 8 X 10-4 m2. What force must be applied to the piston to raise oil (d = 780 Kg/m2) to a height of 6.0 m? Assume the upper end of the oil is open to the atmosphere. 2. A swimmer whose body’s surface area is approximately 1.6 m2 lies at a depth of 3 m below the water surface. How much force is exerted on his body due to water pressure. 3. A hydraulic lift has a narrow cylinder of area 19.64 cm2 and wide cylinder of area 1256.8 cm. Calculate the force that must be applied to the liquid in the small cylinder to lift a car 1950 Kg. 4. Estimate the force exerted on your eardrums due to water when you are swimming at the bottom of a pool that is 5 m deep. 2.7 Buoyant Forces and Archimedes Principles When an object is submerged in a fluid, it appears to weigh less than they do when outside the fluid. For example, a large rock at the bottom of a stream would be easily lifted compare to lifting it from the ground. As the rock breaks through the surface of the water, it becomes heavier. This phenomenon is as a result of upward force called the buoyant force (upthrust) acting on the rock plus the downward gravitational force. The buoyant force occurs because the pressure in fluid increases with depth. Thus, the upward pressure at the bottom surface of a submerged object is greater than the downward pressure at its top surface. Hence, a buoyant force is the upward force exerted by water on any immersed object. Archimedes principle states that a body fully or partially immersed in a fluid experience an upward force (buoyancy) which is equal to the weight of the liquid displaced. Any immersed body will experience two forces the real weight acting vertically downwards and the buoyant force acting vertically upwards. The net downward force is equal to the real weight minus the buoyant force and this is referred to as the apparent 15 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 weight of the body, for a solid object of density 𝝆𝝆𝑺𝑺 and volume V which is fully immersed in a liquid of density 𝝆𝝆𝒍𝒍 and real weight of the solid is 𝝆𝝆𝑺𝑺 𝑽𝑽𝑲𝑲(downwards) and the buoyant force Figure 2.3: Body immersed in a fluid The net downward force or the apparent weight of the object is thus equal to: Apparent weight = 𝝆𝝆𝑺𝑺 𝑽𝑽𝑲𝑲 − 𝝆𝝆𝒍𝒍 𝑽𝑽𝑲𝑲 = (𝝆𝝆𝑺𝑺 − 𝝆𝝆𝒍𝒍 )𝑽𝑽𝑲𝑲 𝟐𝟐. 𝟓𝟓 If the solid is denser than the liquid (i.e. 𝝆𝝆𝑺𝑺 = 𝝆𝝆𝒍𝒍 ) the solid experience a net downward force( a positive apparent weight or a net upward force and it therefore rises. The solid rises in the liquid until the weight of the liquid displaced becomes equal to the real weight of the solid. (i.e. the solid floats partly submerged) 2.7 Surface Tension and Capacity The molecules of a liquid are held together by intermolecular forces. Because of this, a liquid surface acts like a stretched elastic membrane, as if it is under tension. It is for this reason that a steel razor blade, for instance, can be floated on water surface, even though the density of steel is much higher than that of water. The surface forces which holds the molecules near the surface of a liquid together are described by the term surface tension (𝒀𝒀) is defined as the surface force per unit length along a direction 16 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 perpendicular to the force. Thus if a surface force (F) acts perpendicular to a line of length (𝑳𝑳) then, 𝑭𝑭 𝒀𝒀 = 𝟐𝟐. 𝟔𝟔 𝑳𝑳 The magnitude of 𝒀𝒀 varies from one liquid to another, and for a given liquid exert on each other are called cohesive forces. The molecules which are in contact with the wall. This force of attraction is termed adhesion. A liquid surface (which is also called a meniscus) may curve upwards or downwards at the point of contact with the container wall, depending on which of the two force dominant and the angle between the container wall and the meniscus at the point of contact angle (Figure 2.4). When adhesive forces outweigh cohesive forces 𝜽𝜽 < 𝟗𝟗𝟎𝟎𝟎𝟎 (meniscus curves downwards) and when cohesive forces are dominant 𝜽𝜽 > 𝟗𝟗𝟎𝟎𝒐𝒐 meniscus meniscus θ θ θ< 90 o θ> 90 o Figure 2.4: cohesion and adhesion If a very thin tube is inserted into a liquid reservoir, there is a difference between the levels of the liquid inside and outside the tube 𝜽𝜽< 900, the liquid rises higher inside the tube (e.g water in a glass tube) while for 𝜽𝜽> 900 the liquid level is depressed in the tube (as in mercury in glass tube). Consider a liquid which rises to a height h inside a tube of radius r (Figure 5). The cylindrical column of liquid experiences two counter – balancing forces; (1) the upward force F due to the surface tension which acts around the circle of contact between the meniscus and the glass wall 17 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 (2) the downward pull of gravity, or the weight (W) of the liquid column. The balancing of these forces at equilibrium yields. 𝟐𝟐𝟐𝟐 𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽 𝒉𝒉 = 𝟐𝟐. 𝟕𝟕 𝝆𝝆𝑲𝑲𝑺𝑺 for a liquid of surface tension 𝟐𝟐 and density 𝝆𝝆. From equation above, it is clear that h varies inversely as the radius of r of the tube, i.e. a liquid will rise higher in a narrow small cross-section, in which the rise of liquid is quite noticeable, are referred to as capillary tubes and the phenomenon is known as capillarity. Figure 5: display of mercury and water in a test tube Solved Problems 1. A solid aluminum cylinder has a measured mass of 67 g in air and 45 g when immersed in turpentine. What is the buoyant force? Solution: Mass in air = 67 g = 0.067 Kg, Then Weight in air = 0.067 X 9.8 = 0.657 N Mass in turpentine = 45 g = 0.045 Kg, Then Weight = 0.045 X 9.8 = 0.441 N Recall that Buoyant force = Weight in air - Apparent weight in turpentine = 0.657 – 0.441 = 0.216N 2. The xylem tubes which transport sap to the top of a tree can be considered as uniform cylinders. If the transport of sap is entirely due to capillarity, determine the diameter 18 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 of the tubes which will move sap up a tree which is 25 m tall. (Take the specific gravity and surface tension of sap as 1.0 and 5 X 10-2 N/m and contact angle with the tubes as 450 Solution: ℎ = 25 𝑐𝑐 𝛾𝛾 = 5 X 10-2 N/m s.g = 1, then 𝜌𝜌 = 1000 Kg/m3 𝜃𝜃 = 450, r=? 2𝛾𝛾 cos 𝜃𝜃 Using, ℎ = 𝜌𝜌𝜌𝜌𝜌𝜌 2𝛾𝛾 cos 𝜃𝜃 2 (5 𝑋𝑋 10−2 ) cos 45 𝑺𝑺 = = = 𝟐𝟐. 𝟒𝟒𝟗𝟗 𝑿𝑿 𝟏𝟏𝟎𝟎−𝟕𝟕 𝒎𝒎 𝜌𝜌𝐾𝐾ℎ (1000)(9.8)(25) Supplementary Problems 1. A piece of alloy has a measured mass of 86 g in air and 73 g when immersed in water. Find its volume and its density. 2. Two (glass) capillary tubes of diameters 0.05 mm and 2.00 mm are dipped in a pool of water. How high will the water rise in each of the tubes? (Taking contact angle between glass and water is 00) 19 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 MODULE THREE TEMPERATURE AND HEAT Temperature is the measure of the degree of hotness or coldness of a body. Temperature is directly related to motion of molecules making of the body. A rise in temperature of a body is due to increase in the average speed of the molecules (in the case of liquids and gases) or increase in the vibration of molecules about their mean positions (solids). Temperature can therefore be referred to as the average kinetic energy of a molecule in a body or internal energy within a body. A device used in taking the measurement of temperature is known as thermometer. Some measurable properties of matter which includes volume, pressure and electrical resistance easily change with temperature, which then serves as a basis for the three main types of thermometer namely i. Constant volume gas thermometer ii. Mercury in glass thermometer iii. Resistance thermometer. Mercury in glass thermometers are not used for very accurate measurement of temperature due to relatively small range of temperature that can be measured. The freezing point of mercury is -390C and its boiling point is 3600C. Glass also expands and its expansion is irregular. Gas thermometers are used for very accurate temperature measurement. A large volume change of gas occurs when its temperature is altered, so that glass expansion is negligible. Here is an illustration for using thermometer to calculate temperature: suppose X0 is the uncalibrated temperature at ice point 00C; X100 is the uncalibrated temperature at steam point 1000C and XL is the temperature in liquid at T0C. 𝑿𝑿𝑳𝑳 −𝑿𝑿𝟎𝟎 𝑻𝑻𝟎𝟎 𝑪𝑪 =. 1000C (3.1) 𝑿𝑿𝟏𝟏𝟎𝟎𝟎𝟎 −𝑿𝑿𝟎𝟎 20 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 G C Water D Bath Air A Heat Figure 3.1 A simple gas thermometer The two scales used in measuring temperature. There are i. Fahrenheit scale ii. Celsius (or centigrade scale) The temperature at which ice will melt is 320F on the Fahrenheit scale and 00C on the Celsius scale. Water will boil at 2120F on Fahrenheit scale and 1000C on Celsius. The temperature TF on Fahrenheit scale can be converted to the Celsius scale (and vice- versa) using the formula 𝟓𝟓 𝑻𝑻𝑪𝑪 = (𝑻𝑻𝑭𝑭 − 𝟑𝟑𝟐𝟐) (3.2) 𝟗𝟗 Absolute temperature scale (in Kelvin) is one in which the lowest possible is zero (0K). A given Celsius temperature (TC) can be expressed on absolute scale (TK) by using the relation 𝑻𝑻𝑲𝑲 = 𝑻𝑻𝑪𝑪 + 𝟐𝟐𝟕𝟕𝟑𝟑. 𝟏𝟏𝟓𝟓 (𝟑𝟑. 𝟑𝟑) 3.1 Thermal Expansion of Solids The word thermal has to do with heat while expansion means increase in size. In thermal expansion, materials or substance expand or contract on one major factor which is temperature. Thus, solids expand when heated and contract when cooled. 21 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 Thermal expansion plays an important role in numerous applications. For example, thermal expansion joints must be included in buildings, concrete highways and bridges to compensate for changes in dimensions with temperature variations. 3.1.1 Linear Expansion in Solid (Metal) Linear expansion occurs in solid because of the rigidity and the intermolecular force of attraction of the solid. In general, linear expansivity of a solid can be define as the increase in length (𝑙𝑙2 − 𝑙𝑙1 ) per unit length (𝑙𝑙1 ) per unit degree rise in temperature (𝑇𝑇2 − 𝑇𝑇1 ). Fundamentally, linear expansivity is denoted by the symbol alpha (∝) and it is measured in per degree Celsius (℃−1 ) or per degree Kelvin ((°𝐾𝐾 −1 ). Mathematically, ∆𝑙𝑙 ∝= (3.4) 𝑙𝑙1 ∆𝑇𝑇 Let ∆𝑙𝑙 = 𝑙𝑙2 − 𝑙𝑙1 ∆𝑇𝑇 = 𝑇𝑇2 − 𝑇𝑇1 Substituting ∆𝑙𝑙 and ∆𝑇𝑇 into eq. 3.4, we have 𝑙𝑙2 −𝑙𝑙1 ∝= (3.5) 𝑙𝑙1 (𝑇𝑇2 −𝑇𝑇1 ) From equation 3.5, 𝑙𝑙2 = 𝑙𝑙1 (1+∝ ∆𝑇𝑇) (3.6) Where ∝ = Linear expansivity 𝑙𝑙2 = Length of metal at temperature 𝑇𝑇2 𝑙𝑙1 = Length of metal at temperature 𝑇𝑇1 ∆𝑇𝑇 = Change in temperature 22 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 WORKED EXAMPLE 1. A steel rod increases its length by 5 𝑐𝑐𝑐𝑐 when the temperature increases by 10℃. What is the initial length of the rod if the coefficient of linear expansion for steel is 1.1 × 10−5 per 0C? Solution: Recall that ∆𝐿𝐿 ∝= 𝐿𝐿1 ∆𝑇𝑇 ∆𝐿𝐿 5 X 10−3 𝐿𝐿1 = = = 𝟒𝟒𝟓𝟓. 𝟒𝟒𝟓𝟓𝟒𝟒 ∝ ∆𝑇𝑇 1.1 X 10−5 x10 2. Steel rails 8.0 m long are laid end to end in winter when the temperature is −10℃. (i) How much space should be left between them to allow for expansion in summer when the temperature could reach 50℃? (ii) If the rails had been butted against each other at −10℃, what stress would each have to withstand at 50℃ in order to prevent buckling? (Take ∝= 1.2 × 10−5 ℃−1 and 𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝐾𝐾𝑠𝑠 𝑀𝑀𝑌𝑌𝑚𝑚𝑌𝑌𝑙𝑙𝑌𝑌𝑠𝑠 (𝐸𝐸) = 21 × 1010 𝑁𝑁⁄𝑐𝑐2 ) Solution (a) Recall that ∆𝑙𝑙 ∝= 𝑙𝑙1 ∆𝑇𝑇 ∆𝑙𝑙 =∝ 𝑙𝑙1 (𝑇𝑇2 − 𝑇𝑇1 ) ∆𝑙𝑙 = 1.2 × 10−5 ℃−1 × 8.0 m [50℃ − (−10℃)] ∆𝒍𝒍 = 𝟓𝟓. 𝟕𝟕𝟔𝟔 × 𝟏𝟏𝟎𝟎−𝟑𝟑 𝒎𝒎 (b) Recall from previous knowledge that 𝑆𝑆𝐷𝐷𝐷𝐷𝐷𝐷𝑠𝑠𝑠𝑠 (𝜎𝜎) = 𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝐾𝐾𝑠𝑠 𝑐𝑐𝑌𝑌𝑚𝑚𝑌𝑌𝑙𝑙𝑌𝑌𝑠𝑠 (𝐸𝐸) × 𝑆𝑆𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝑌𝑌(∈) 10 2 5.76 × 10−3 𝑐𝑐 𝑆𝑆𝐷𝐷𝐷𝐷𝐷𝐷𝑠𝑠𝑠𝑠 (𝜎𝜎) = 21 × 10 𝑁𝑁⁄𝑐𝑐 × 8.0 𝑐𝑐 23 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 𝑺𝑺𝑺𝑺𝑺𝑺𝒆𝒆𝑺𝑺𝑺𝑺 (𝝈𝝈) = 𝟏𝟏. 𝟓𝟓𝟏𝟏 × 𝟏𝟏𝟎𝟎𝟒𝟒 𝑵𝑵⁄𝒎𝒎𝟐𝟐 3.1.2 Area Expansion It has been observed that the area of a metal changes when heated. Thus, area expansion has a relationship with coefficient of superficial expansion (𝛽𝛽), original area (𝐴𝐴1 ) and its temperature difference (∆𝑇𝑇). We therefore, define coefficient of superficial expansion as the increase in area (∆𝐴𝐴) per unit area (𝐴𝐴1 ) per unit degree rise in temperature (∆𝑇𝑇). Mathematically, ∆𝑨𝑨 𝜷𝜷 = (3.7) 𝑨𝑨𝟏𝟏 ∆𝑻𝑻 Let ∆𝐴𝐴 = 𝐴𝐴2 − 𝐴𝐴1 ∆𝑇𝑇 = 𝑇𝑇2 − 𝑇𝑇1 Putting ∆𝐴𝐴 and ∆𝑇𝑇 in equation 3.7, we have 𝐴𝐴2 −𝐴𝐴1 𝛽𝛽 = (3.8) 𝐴𝐴1 (𝑇𝑇2 −𝑇𝑇1 ) From equation 3.8, 𝐴𝐴2 = 𝐴𝐴1 (1 + 𝛽𝛽∆𝑇𝑇) (3.9) Where 𝛽𝛽 = Superficial or area expansivity of the metal 𝐴𝐴2 = Length of metal at temperature 𝑇𝑇2 𝐴𝐴1 = Length of metal at temperature 𝑇𝑇1 ∆𝑇𝑇 = Change in temperature Research has shown that the coefficient of superficial expansion has a relationship with linear expansivity as given below 𝜷𝜷 = 𝟐𝟐 ∝ (3.9.1) 24 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 Worked Example The coefficient of linear expansivity of a metal is 0.000019 °𝐾𝐾 −1. What will the area of 400 mm2 of the metal become if its temperature is raised by 10 K? Solution: Recall that 𝐴𝐴2 = 𝐴𝐴1 (1 + 𝛽𝛽∆𝑇𝑇) But, 𝛽𝛽 = 2 ∝ Therefore, 𝐴𝐴2 = 𝐴𝐴1 (1 + 2 ∝ ∆𝑇𝑇) 𝐴𝐴2 = 400 mm2 (1 + 2(0.000019 °𝐾𝐾 −1 ) × 10 K) 𝑨𝑨𝟐𝟐 = 𝟒𝟒𝟎𝟎𝟎𝟎. 𝟏𝟏𝟓𝟓𝟐𝟐 𝟒𝟒𝟒𝟒𝟐𝟐 3.2 Cubical Expansivity of a Liquid Liquids are always held in vessels and a study of the expansion of liquids is often complicated by the expansion of its container. Liquids have no length or surface area of their own but always take the shapes of their containers. We therefore never talk of the linear expansivity of a liquid, but its cubical expansivity relative to the material of the container. To this end, we will distinguish between the real and apparent expansion of the liquid since the expansion of the real vessel complicates the expansion of the liquid under study. 3.2.1 Real Cubical Expansivity of Liquid This is defined as the increase in volume per unit volume per unit rise in temperature. Mathematically, it can be written as ∆𝑽𝑽 𝟐𝟐𝑺𝑺 = (3.9.2) 𝑽𝑽𝟏𝟏 ∆𝑻𝑻 𝑽𝑽𝟐𝟐 −𝑽𝑽𝟏𝟏 𝟐𝟐𝑺𝑺 = (3.9.3) 𝑽𝑽𝟏𝟏 (𝑻𝑻𝟐𝟐 −𝑻𝑻𝟏𝟏 ) 3.2.2 The Apparent Cubical Expansivity of a Liquid This is define as the increase in volume per unit volume per unit rise in temperature when the liquid is heated in an expansible vessel and is denoted by 𝛾𝛾𝑎𝑎. Therefore, 25 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 𝟐𝟐𝑺𝑺 = 𝟐𝟐𝑺𝑺 + 𝟐𝟐 (3.9.3) Where 𝛾𝛾𝜌𝜌 = real cubical expansivity 𝛾𝛾𝑎𝑎 = apparent cubical expansivity 𝛾𝛾= Cubic expansivity of the vessel N:B 𝟐𝟐= 3∝ Worked Example A glass flask of volume 100 𝑐𝑐𝑐𝑐3 is filled to the brim with liquid whose cubical coefficient of expansion is 1 × 10−3 ℃−1. The flask and its liquid content are originally at 20℃. Determine the volume of liquid which will overflow upon heating the flask and the liquid to 50℃ if the coefficient of linear expansion of glass is 8 × 10−6 ℃−1. Solution: Recall that ∆𝑉𝑉 𝛾𝛾𝜌𝜌 = 𝑉𝑉1 ∆𝑇𝑇 Therefore, for liquid ∆𝑉𝑉 = 𝛾𝛾𝐿𝐿 𝑉𝑉1 ∆𝑇𝑇 ∆𝑉𝑉 = 1 × 10−3 ℃−1 × 100 𝑐𝑐𝑐𝑐3 (50℃ − 20℃) ∆𝑉𝑉 =× 3.0 𝑐𝑐𝑐𝑐3 Also for glass, ∆𝑉𝑉 = 𝛾𝛾𝜌𝜌 𝑉𝑉1 ∆𝑇𝑇 But 𝛾𝛾𝜌𝜌 = 3∝𝜌𝜌 Therefore, ∆𝑉𝑉 = 3 ∝𝜌𝜌 𝑉𝑉1 ∆𝑇𝑇 ∆𝑉𝑉 = 3 × 8 × 10−6 ℃−1 × 100 𝑐𝑐𝑐𝑐3 (50℃ − 20℃) ∆𝑽𝑽 = 𝟎𝟎. 𝟎𝟎𝟕𝟕𝟐𝟐 𝒔𝒔𝒎𝒎𝟑𝟑 Hence, the volume of liquid which overflows 26 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 = 3.0 𝑐𝑐𝑐𝑐3 − 0.072 𝑐𝑐𝑐𝑐3 = 𝟐𝟐. 𝟗𝟗𝟐𝟐𝟒𝟒 𝒔𝒔𝒎𝒎𝟑𝟑 3.3 Heat Capacity Heat is a form of energy and transfer of heat occurs when two bodies with different temperatures and are in thermal contact. The unit of heat is Joules. A body does not contain heat but internal energy. The quantity heat required to raise the temperature of material by 10C is known as heat capacity. Heat capacity varies from material to material. The quantity of heat required to raise the temperature of a unit mass of a material by 10C is specific heat capacity of that material. Thus to raise a mass m of the substance by ΔT, we need a quantity of heat ΔQ given by ∆𝑸𝑸 = 𝒎𝒎𝒔𝒔∆𝑻𝑻 (3.9.4) The molar specific heat C of a substance is the heat required to raise the temperature of 1 mole of the substance by 10C. For n moles of the substance ∆𝑸𝑸 = 𝑺𝑺𝑪𝑪∆𝑻𝑻 (3.9.5) The number of moles (n) in a substance of mass m is 𝒎𝒎 𝑺𝑺 = (3.9.6) 𝑴𝑴 where, M is the molecular weight of the substance 3.4 Matter and Phase Change Matter can exist in solid, liquid and gaseous phases. The particular phase in which a substance will be found depends on the temperature and pressure. Changes from one phase to another require the addition or removal of heat (energy), even though such changes usually occur at fixed temperatures. The heat absorbed or released during a phase change is called latent heat (L). The latent heat of fusion is the heat required to melt a unit substance while the latent heat of vaporization, (Lv)is the heat required 27 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 to convert a unit mass of a substance from liquid to vapour. The heat required to change the phase of mass m of a substance is thus 𝑸𝑸𝑭𝑭 = 𝒎𝒎𝑳𝑳𝑭𝑭 (for fusion) (3.9.7) and 𝑸𝑸𝑽𝑽 = 𝒎𝒎𝑳𝑳𝑽𝑽 (for vaporization) (3.9.8) The units of LF and LV are J/kg Solved Problems 1. In an uncalibrated mercury thermometer, the length of the mercury thread above the bulb, in the capillary, is 18mm at a temperature of melting ice and 138mm at a temperature of steam. When placed in hot liquid, the length of the mercury thread is is 118mm. Calculate the temperature of the liquid. Solution Using Eq. 4.1 118𝑚𝑚𝑚𝑚−18𝑚𝑚𝑚𝑚 𝑻𝑻𝟎𝟎 𝑪𝑪 = 𝑥𝑥 100°𝐶𝐶 = 830C 138𝑚𝑚𝑚𝑚−18𝑚𝑚𝑚𝑚 2. At what temperature will the Celsius and Fahrenheit temperature scales record the same reading? Solution Using Eq. 4.2 5 𝑇𝑇𝐶𝐶 = (𝑇𝑇𝐹𝐹 − 32) 9 Let TC = TF = T, so that 5 𝑇𝑇 = (𝑇𝑇 − 32) 9 T = -400C or -400F 3. A malaria patient has a body temperature of 39.50C. Convert this temperature to (a) 0F (b) K Solution 28 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 9 9 𝑻𝑻𝑭𝑭 = 𝑇𝑇𝐶𝐶 + 32 = (39.5) + 32 5 5 TF = 103.10C 4. A steel rod increases its length by 5mm when the temperature increases by 100C. What is the initial length of the rod if the coefficient of linear expansion for steel is 1.1x10-5per 0C? Solution: ∆𝐿𝐿 ∝= 𝐿𝐿∆𝑇𝑇 ∆𝐿𝐿 5 X 10−3 𝐿𝐿 = = = 𝟒𝟒𝟓𝟓. 𝟒𝟒𝟓𝟓𝟒𝟒 ∝ ∆𝑇𝑇 1.1 X 10−5 x10 5. How much heat must be added to a 4.0x10-3kg steel ball bearing in order to increase its temperature by 30K if the specific heat capacity of steel is 4.49x102J/Kg.K? And how much will the temperature of the ball increase if it were made of gold of specific heat capacity 1.29x102J/Kg.K rather than steel? Solution 𝑄𝑄 = 𝑐𝑐𝑐𝑐∆𝑇𝑇 = (4.0x10-3)( 4.49x102)(30) = 53.88 J For gold 𝑄𝑄 53.88 ∆𝑇𝑇 = = = 𝟏𝟏𝟎𝟎𝟒𝟒. 𝟒𝟒𝟒𝟒 𝑚𝑚𝑚𝑚 (4 X 10−3 )(1.29 X 102 ) A 2.0x10-2kg ice at 00C is dropped into a vacuum bottle originally holding 0.4kg of water at 350C. Assuming that any loss or gain of heat by the vacuum bottle is negligible, determine (a) the heat to melt the ice and (b) the final temperature after thermal equilibrium is attained Solution (a) Q = mLF = (2x10-2) (3.35x105) = 6.7x103J 29 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 (b) Let the final temperature be T. Heat lost by water originally at 350C = heat required to melt ice + heat required to warm resulting ice-water to T (0.4) (4.18x103) (35-T) = 6.7x103 + (0.02) (4.18x103)T T = 29.60C Supplementary Problems 1. What is the absolute temperature of boiling water and melting ice? 2. Determine the temperature whose Fahrenheit and Kelvin scales have the same reading. 3. A scientist uses a γ scale for measuring temperature. In this scale water melts at 100γ, and boils at 1300γ. The scientist measures the temperature at which sodium melts to be 1270γ. Express this temperature in Kelvin. 4. A glass flask of volume 100cm3 is filled to the brim with liquid whose cubical coefficient of expansion is 1x10-3 per 0C. The flask and its liquid content are originally at 200C. Determine the volume of liquid which will overflow upon heating the flask to 500C if the coefficient of linear expansion of glass is 8x10-6 per 0C? 5. 500cm3 of water is to be heated from room temperature (280C) to 1000C in order to prepare hot cup of coffee. (a) What is the minimum heat required? (b) How long will it take to heat the water with a 1000W heating coil which has a heating efficiency of 70%? 6. A 40kg metal slab at temperature 6000C is taken from a furnace and plunged into 300kg of oil originally at 250C. The final temperature of the oil/slab is 400C. Determine the specific heat capacity of the metal if that of oil is 2100J/kg.0C. 7. A 100g ice-block at -200C is dumped into a thermally insulated container of water at 00C. How much water is frozen if the specific heat of ice is 2302 J/Kg.0C and latent heat of fusion is 3.35x105 J/Kg? 30 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 MODULE FOUR HEAT TRANSFER If two bodies at initially different temperature are brought into thermal contact, thermal equilibrium can be attained if heat is able to flow from the hotter body to the cooler body. Conduction, convection and radiation are the three main processes by which heat is transferred between two regions of different temperature. 4.1 CONDUCTION Conduction is the transfer of heat energy from one region of a body to another region without the actual movement of any part of the body. An example of conduction is the transfer of heat energy from the hot end of a metal rod placed in fire to the cold 𝒅𝒅𝑸𝑸𝑩𝑩 end of the rod which feels warmer and warmer. The rate of heat flow, by conduction 𝒅𝒅𝑺𝑺 is given by 𝒅𝒅𝑸𝑸𝑩𝑩 𝒅𝒅𝑻𝑻 = −𝑩𝑩𝑨𝑨𝒔𝒔 (4.1) 𝒅𝒅𝑺𝑺 𝒅𝒅𝒙𝒙 where, K is the thermal conductivity of the rod material, Ac is cross-sectional area of 𝒅𝒅𝑻𝑻 rod and is the temperature gradient. The temperature gradient is negative since T 𝒅𝒅𝒙𝒙 decreases as x increases. Figure 4.1 Conduction in a metal rod. 31 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 Eq. 4.1 becomes 𝒅𝒅𝑸𝑸𝑩𝑩 ∆𝑻𝑻 𝑻𝑻𝑪𝑪 −𝑻𝑻𝑯𝑯 = −𝑩𝑩𝑨𝑨𝒔𝒔 = −𝑩𝑩𝑨𝑨𝒔𝒔 (4.2) 𝒅𝒅𝑺𝑺 ∆𝑳𝑳 𝑳𝑳 4.1.1 Conductors and insulators Materials, for example metal that have high values of k are good conductors of heat. Insulators which are mostly nonmetals are poor heat conductors with low values of k. Conduction is a result of vibration of molecules upon increase in temperature. The vibrating molecules set other neighboring molecules into vibration after colliding with them. The vibratory motion is eventually transmitted from the hot end of the metal rod to the colder end. 4.2 CONVECTION It is the transfer of heat from one region to another by the actual motion of the heated medium. This process of heat transfer finds application only in liquid and gases. 𝒅𝒅𝑸𝑸𝒔𝒔 The rate of heat flow by convection can be expressed as 𝒅𝒅𝑺𝑺 𝒅𝒅𝑸𝑸𝒔𝒔 = 𝒉𝒉𝑨𝑨𝑺𝑺 (𝑻𝑻𝑺𝑺 − 𝑻𝑻𝑭𝑭 ) (4.3) 𝒅𝒅𝑺𝑺 h is convection coefficient, As is area of contact between surface and fluid, Tf is temperature of the main body of the fluid and TF is the temperature of the surface. Figure 4.3: Heat convection from the surface to the fluid (liquid or gas). 32 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 4.3 RADIATION Radiation is the transfer of heat energy between two bodies that are not in physical contact. It involves the transfer of heat through space. The earth is heated by the sun through the radiation process. All surfaces emit energy in form of electromagnetic wave (i.e infrared) and electromagnetic wave is converted to heat energy the instant it is absorbed by a body. The intensity of radiant it depends on the nature and temperature of the surface. For a surface of area A at absolute temperature T1, the rate at which radiant heat is emitted is given as 𝒅𝒅𝑸𝑸𝑹𝑹,𝑬𝑬 = 𝜺𝜺𝝈𝝈𝑨𝑨𝑻𝑻𝟒𝟒𝟏𝟏 (4.4) 𝒅𝒅𝑺𝑺 Where σ is the Stefan-Boltzmann constant which has a value of 5.6703 x 10- 8 Watt/m2.K4. ε is the emissivity and lies between 0 and 1 depending on the nature of the surface. The rate at which a body absorbs radiant energy when completely surrounded by an enclosure whose temperature is maintained at T2is 𝒅𝒅𝑸𝑸𝑹𝑹,𝑨𝑨 = 𝜺𝜺𝝈𝝈𝑨𝑨𝑻𝑻𝟒𝟒𝟐𝟐 (4.5) 𝒅𝒅𝑺𝑺 The net rate of radiant heat flow from a body at absolute temperature T1 which is surrounded by an enclosure whose surface is at absolute temperature T2 is 𝒅𝒅𝑸𝑸𝑹𝑹,𝑨𝑨 = 𝜺𝜺𝝈𝝈𝑨𝑨(𝑻𝑻𝟒𝟒𝟏𝟏 − 𝑻𝑻𝟒𝟒𝟐𝟐 ) (4.7) 𝒅𝒅𝑺𝑺 33 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 Figure 5.7: Radiant heat in an enclosure A surface that absorbs the entire radiation incident upon it would appear black since no radiation would be reflected from it. Such body is an ideal absorber and ideal emitter with ε = 1. It is called a black surface and such body is referred to as blackbody. Solved Problems 1. Determine the quantity of heat which is conducted in 30 minutes through an iron plate 2.0cm thick and 0.10 m2 in area if the temperature of the two sides 00C and 200C. The coefficient of thermal conductivity of iron is 0.12ca/s.cm.C0. Solution AC = 0.1m2, ΔT = -200C, ΔL = 2 x 10-2m Convert k in cal to Joules k = 0.12 x (4.184) J/s.m.C0 = 50.4 J/s.m.C0 𝑑𝑑𝑄𝑄𝑘𝑘 ∆𝑇𝑇 −50.4 (0.1)(−20) Using Eq. 6.2 = −𝑘𝑘𝐴𝐴𝑚𝑚 = (2 𝑥𝑥 10−2 ) = 𝟓𝟓𝟎𝟎𝟒𝟒𝟎𝟎 𝐉𝐉/𝐜𝐜 𝑑𝑑𝑑𝑑 ∆𝐿𝐿 Heat conducted in 30 minutes = 5040 x 60 x 30 = 9.07 x 10-6 J. 34 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 2. The inside surface of a wall of a home is maintained at constant temperature of 250C while the inside air is at 150C. How much heat is lost by natural convection from the 8.0m x 4.0 m in 24 hours if the average convection coefficient is 3.49 J/s.m2C0? Solution A = 8.0 x 4.0 = 32 m2 ΔT = 25 – 15 = 100C h = 3.49 J/s.m2C0 Using Eq. 6.4 𝑑𝑑𝑄𝑄𝑐𝑐 = ℎ𝐴𝐴𝐴𝐴𝑇𝑇 = (3.48 x 32 x 10) = 1116.8 J/s 𝑑𝑑𝑑𝑑 In 24 hours, heat lost = 1116.8 x 24 x 3600 = 9.65 x 10-7 J (a) Determine the rate of radiant emission per m2 from a blackbody at (a) 250C, (b) 30000C Solution A = 1 m2 ε=1 (a) T =273 + 25 =298 K 𝑑𝑑𝑄𝑄𝑅𝑅,𝐸𝐸 = 𝜀𝜀𝜎𝜎𝐴𝐴𝑇𝑇 4 = 1.0 x (5.67 x 10-8) x 1 x (298)4 = 395.5 W 𝑑𝑑𝑑𝑑 (b) T = 273 + 3000 = 3273 K 𝑑𝑑𝑄𝑄𝑅𝑅,𝐸𝐸 = 𝜀𝜀𝜎𝜎𝐴𝐴𝑇𝑇 4 = 1.0 x (5.67 x 10-8) x 1 x (3273)4 = 6507 kW 𝑑𝑑𝑑𝑑 Supplementary Problems 1. One end of a 30 cm long aluminum rod is exposed to a temperature of 5000C while the other end is maintained at 200C the rod has a diameter of 2.5 cm. if heat is conducted through the rod at the rate of 142 kcal/hr, calculate the thermal conductivity of aluminum. 35 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 2. A thin hot plate which measures 20 cm x 20 cm is maintained at a temperature of 1000C. It is suspended in air at 250C. Determine the heat transferred by convection from both sides of the plate to the surrounding air in one hour if the coefficient of convective heat transfer is 5.0 J/s.m2.C0. 3. Determine the electric power that must be supplied to the filament of a bulb operating at 3000K. The total surface area of the filament is 8 x 10-6 m2 and its emissivity is 0.92. 4. A 10 cm diameter metal sphere of emissivity 0.9 is located in a room whose walls are maintained at a temperature 270C. At what rate must energy be supplied to the sphere in order to maintain its temperature at 1000C? (neglected heat loss by convection) 36 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 MODULE FIVE LAWS OF THERMODYNAMICS Thermodynamics is the study of relationship between heat and other forms of energy. Heat can do work and work can generate heat. An example of how heat can do work is: the mechanical work done by the engine of an automobile is a consequence of the fuel burnt (heat). A simple description of how work can generate heat is rubbing of palms together (mechanical work) to produce heat. So thermodynamics is work and heat exchanges of matter. Thermodynamic system refers to the matter under study in relation to its immediate surroundings which is affected by heat changes within the system. A closed thermodynamic system is that in which the quantity of matter within the region under investigation is constant. An open system is one which is free to exchange matter and heat with its surroundings. An isolated system is one which has net exchange of matter and heat with its surroundings. 4.1 Zeroth Law of Thermodynamics Heat is a thermal energy that flows from one body or system to another, which is in contact with it, because of their temperature difference. Heat flow from hot to cold. For two objects in contact to be in thermal equilibrium witheach other ( i.e. for no net heat transfer from one another), their temperatures must be the same. If each of two objects is in thermal equilibrium with a third body, then the two objects are in thermal equilibrium with each other. This fact is often referred to as the Zeroth Law of Thermodynamics 4.2 First Law of Thermodynamics It states that if an amount of heat ΔQ flows into a system, then this energy must appear as increased internal energy ΔU for the system and work ΔW done by the system 37 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 on its surroundings. The first law is a statement of the law of conservation of energy. The law can be expressed as ∆𝑸𝑸 = ∆𝑼𝑼 + ∆𝑾𝑾 (5.11) 4.2.1 Internal energy (ΔU) The internal energy of a system is the total energy content of the system. It is the sum of all forms of energy possessed by the atoms and molecules of the system. 4.2.2 Work done by a system (ΔW) ΔW is positive if the system thereby loses energy to its surroundings. When the surroundings do work on the system so as to give it energy, ΔW is a negative quantity. In a small expansion ΔV, a fluid at constant pressure P does work given by ∆𝑾𝑾 = 𝑷𝑷∆𝑽𝑽 = 𝑷𝑷(𝑽𝑽𝟐𝟐 − 𝑽𝑽𝟏𝟏 ) (5.12) 4.2.3 Thermodynamic processes Thermodynamic process is a process where there are changes in the state of the thermodynamic system. 4.2.3a Isobaric process It is a process carried out under constant pressure. 4.2.3b Isochoric process It is a process carried out under constant volume. When a gas undergoes such process, ∆𝑾𝑾 = 𝑷𝑷∆𝑽𝑽 = 𝟎𝟎 (5.13) So the first law of thermodynamics becomes ∆𝑸𝑸 = ∆𝑼𝑼 (5.14) Any heat that flows into the system appears as increased energy of the system. 38 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 4.2.3c Isothermal process It is the process done under constant temperature. In the case of an ideal gas where the constituent atoms or molecules do not interact, ΔU = 0 in an isothermal process. However, this is not true for many other systems. For example, ∆𝑈𝑈 ≠ 0, as ice melts to water at 00C, even though the process is isothermal. For an ideal gas, ΔU = 0 in an isothermal change and so the First Law becomes ∆𝑸𝑸 = ∆𝑾𝑾 (5.15) For an ideal gas changing isothermally where P1V1 = P2V2, 𝑽𝑽 ∆𝑸𝑸 = ∆𝑾𝑾 = 𝑷𝑷𝟏𝟏 𝑽𝑽𝟏𝟏 𝑰𝑰𝑺𝑺( 𝟐𝟐) (5.16) 𝑽𝑽𝟏𝟏 4.2.3d Adiabatic process In adiabatic process, there is no heat transfer between systems. ΔQ = 0 in such process. Hence the first law becomes 0 = ΔU + ΔW (5.17) Any work done in this system increases the internal energy. (ΔU = -ΔW). Important notice:∆𝑸𝑸is positive when heat is added to the system, and ΔW is positive when the system does work. In reverse cases ∆𝑸𝑸and ∆𝑾𝑾 must be taken as negative. 4.3 Entropy and Second Law of Thermodynamics Entropy is a measure of molecular disorder or randomness in a system. It can also be defined as the measure of a system’s thermal energy per unit temperature that is unavailable for any useful work. Entropy is expressed as 39 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 ∆𝑸𝑸 ∆𝑺𝑺 = (5.18) 𝑻𝑻 The second law of thermodynamics states that the entropy of a system can never decrease with time but can remain constant only if such process is reversible. The second law can be clearly stated in two other ways: 1. It is impossible for any system to undergo a process in which it absorbs heat from a reservoir at a single temperature and converts the heat completely into mechanical work with the system ending in the same state in which it began. 2. It is impossible for any process to have as its sole result the transfer of heat from a cooler to a hotter body. The first statement deals with heat engines while the second statement can be viewed by considering refrigerators. 4.3.1 Heat Engines A heat engine is a device which converts heat energy to work. Such engine involves a cyclic process in which initial and final states are identical. In such engine the working substance is usually taken through a cycle process in which the initial and final states are identical. A common heat engine is the petrol engine, the type in which the workdone is used in propelling automobile. The working cycle for the petrol engine consists of four main processes (Fig. 5.1); A to B: Adiabatic compression of air-petrol mixture (the working substance) from atmospheric pressure to state B. (compression stroke) B to C: heating at constant volume to state 3 (combustion). C to D: Adiabatic expansion to state 4 (working stroke) D to A: Cooling at constant volume to state 1 (exhaust) 40 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 Figure 5.1: Working cycle of petrol engine. In this engine, heat QA is added to the system during combustion process while heat QR is rejected to the surroundings during exhaust. The network output of the engine (W) of the engine is 𝑾𝑾 = 𝑷𝑷(𝑽𝑽𝟐𝟐 − 𝑽𝑽𝟏𝟏 ) or the shaded area. For any system taken through a complete cycle, the internal energy at the initial and final states is the same. Applying the first law to the cycle gives 𝑸𝑸 = 𝑾𝑾(the network produced by the engine is equal to the net heat absorbed in a cyclic process). Since the net heat absorbed is equal to the heat added QA minus the heat rejected QR the first law for the cyclic process becomes 𝑸𝑸𝑨𝑨 − 𝑸𝑸𝑹𝑹 = 𝑾𝑾 (5.19) 4.3.1a Thermal efficiency of heat engine It is the ratio of the work done by the engine to heat added. Thermal efficiency η can be expressed as 𝑾𝑾 𝑸𝑸𝑨𝑨 − 𝑸𝑸𝑹𝑹 𝜼𝜼 = = (5.20) 𝑸𝑸𝑨𝑨 𝑸𝑸𝑨𝑨 or 𝑸𝑸𝑹𝑹 𝜼𝜼 = 𝟏𝟏 − (5.21) 𝑸𝑸𝑨𝑨 𝑸𝑸𝑹𝑹 The most efficient engine is thus one for which the ratio is made small as possible. 𝑸𝑸𝑨𝑨 41 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 The Carnot cycle I,s the most efficient cycle possible for a heat engine. An engine that operate according to this cycle between a hot reservoir (TH) and a cold reservoir (TC) has efficiency 𝑻𝑻𝑪𝑪 𝜼𝜼𝑪𝑪 = 𝟏𝟏 − (5.11) 𝑻𝑻𝑯𝑯 4.4 Refrigerators A refrigerator can be described as a heat engine operating in the reverse. It removes heat from a hot place to a colder place. While a heat engine has a net output of mechanical work, a refrigerator has a net input of mechanical work. W and QH are negative in the refrigerator based on the sign conventions earlier described. The first law becomes −𝑄𝑄𝐻𝐻 = 𝑄𝑄𝐶𝐶 − 𝑊𝑊 Where QH is the heat going into the hot surroundings (hot reservoir) and QC is the heat taken from inside the refrigerator (cold reservoir). 5.5 Solved Problems 1. For each of the following Adiabatic processes, find the internal energy (a) A gas does 5J of work while expanding adiabatically. (b) During an adiabatic compression, 80J of work is done on a gas. Solution During adiabatic process, no heat is to and from the system i.e ΔQ = 0 (a) ΔU = 0 – ΔW = 0 – 5 = -5J (b) ΔU = 0 – (- 80) = +80J 42 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 2. A 50 kg mass is placed on a piston fitted to a gas cylinder. If 149J of heat energy is supplied to the gas cylinder, increasing its internal energy by 100J, determine the height to which the mass on the piston is raised. Solution Q = 149J, ΔU = 100J W = Q – ΔU = 149 – 100 = 49J To determine height, W =mgh, therefore, h = W/mg = 0.10m 3. A Carnot engine is operated between two heat reservoirs at temperatures 400K and 300K. If the engine receives 2000 cal from the 400 K reservoir, (a) how many calories does it reject to the lower temperature reservoir? (b) What is the thermal efficiency of the engine? Solution: Applying Eq. 5.1.0 𝑄𝑄ℎ 𝑇𝑇𝐶𝐶 300 (a) 𝑄𝑄𝐶𝐶 = = (2000)( ) = 1500cal. 𝑇𝑇𝐻𝐻 400 𝑇𝑇𝐶𝐶 (b) 𝜂𝜂𝐶𝐶 = 1 − = 1- (300/400) = 0.25 (or 25%) 𝑇𝑇𝐻𝐻 Supplementary Problems 1. In each of the following situations, find the change in the internal energy of the system. (a) A system absorbs 500 cal of heat and at the same time does 420 J of work. (b) A system absorbs 300 cal of heat and at the same time 420 J of work is done on it. (c) 1200 calories is removed from a gas held at constant volume. Give your answers in kilojoules. 2. What is the maximum efficiency of an engine which operates between two reservoirs at temperatures of 250C and 40 oC. 43 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 3. A Carnot engine operating between two reservoirs at temperatures 00C and 1000C receives 10 kJ of heat from the high temperature reservoir. Calculate (a) The heat rejected to the low temperature reservoir. (b) The work done by the engine (c) The thermal efficiency 44 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 MODULE SIX KINETIC MOLECULAR THEORY OF GASSES 6.1 Gas Law For a fixed mass of gas, these three variables can be considered. i.e. the pressure (P), the volume (V) and the absolute temperature (T). If one variable is kept constant, then, the two variables will have some definite relationship. 6.1.1 Boyle’s Law Robert Boyles in 1662, after performing experiments showed that the pressure, P of a fixed mass of a gas is inversely proportional to its volume provided the absolute temperature remains constant. Mathematically, 𝟏𝟏 𝑷𝑷 ∝ (T being constant) (6.1) 𝑽𝑽 𝑲𝑲 𝑷𝑷 = 𝑽𝑽 𝑷𝑷 𝑽𝑽 = 𝑲𝑲 where, 𝐾𝐾 is the constant of proportionality. For any two states: 𝑷𝑷𝟏𝟏 𝑽𝑽𝟏𝟏 = 𝑷𝑷𝟐𝟐 𝑽𝑽𝟐𝟐 (6.2) 45 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 Graphically, V P Figure 6.1: Graphical illustration of Boyle’s law 6.1.2 Charles’s Law It states that the volume of a fixed mass of gas at constant pressure is directly proportional to its absolute temperature. Mathematically, 𝑽𝑽 ∝ 𝑻𝑻 (P being constant) 𝑽𝑽 = 𝑲𝑲𝑻𝑻 𝑽𝑽 = 𝑲𝑲 (6.3) 𝑻𝑻 For any two states: 𝑽𝑽𝟏𝟏 𝑽𝑽𝟐𝟐 = (6.4) 𝑻𝑻𝟏𝟏 𝑻𝑻𝟐𝟐 where, K is Constant of proportionality. 46 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 Graphically, V T Figure 6.2: Graphical illustration of Charles Law 6.1.3 Gay- Lussac’s Law It states that pressure is directly proportional to the absolute temperature provided the volume is fixed. Mathematically, 𝑷𝑷 ∝ 𝑻𝑻 𝑷𝑷 = 𝑲𝑲𝑻𝑻 (6.5) 𝑷𝑷 = 𝑲𝑲 𝑻𝑻 The three relationship in equation 6.1 to 6.5 can be combined to give the equation of state: 𝑷𝑷𝑽𝑽 = 𝒔𝒔𝒐𝒐𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺 (6.6) 𝑻𝑻 For any two states: 𝑽𝑽𝟏𝟏 𝑷𝑷𝟏𝟏 𝑽𝑽𝟐𝟐 𝑷𝑷𝟐𝟐 = 6.7 𝑻𝑻𝟏𝟏 𝑻𝑻𝟐𝟐 47 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 The constant in equation (1.6) has been shown, through experiments as 𝑷𝑷𝑽𝑽 = 𝑺𝑺𝑹𝑹𝑻𝑻 6.8 where n is the number of moles of gas and R is the universal gas constant with the value 8.314 J/mol.K. or 8314KJ/Kmol.K A gas for which equation (6.8) holds is called the idea or a perfect gas and the equation is thus called ideal gas equation. For an ideal gas, 𝑷𝑷𝑽𝑽 = 𝟏𝟏 6.9 𝑺𝑺𝑹𝑹𝑻𝑻 6.2 Dalton’s Law of Partial Pressure Consider a gaseous mixture which consists of 𝑺𝑺𝟐𝟐 , and 𝑺𝑺𝟑𝟑 moles of gases 1, 2 and 3 respectively. It is assumed that the gases do not react with one another: The total number of moles in the mixture is𝑺𝑺 = 𝑺𝑺𝟏𝟏 + 𝑺𝑺𝟐𝟐 + 𝑺𝑺𝟑𝟑. Equation 6.8 can be written for mixture as 𝑷𝑷𝑽𝑽 = (𝑺𝑺𝟏𝟏 + 𝑺𝑺𝟐𝟐 + 𝑺𝑺𝟑𝟑 )𝑹𝑹𝑻𝑻 𝑺𝑺𝟏𝟏 𝑹𝑹𝑻𝑻 𝑺𝑺𝟐𝟐 𝑹𝑹𝑻𝑻 𝑺𝑺𝟑𝟑 𝑹𝑹𝑻𝑻 𝑷𝑷 = + + 𝟔𝟔. 𝟏𝟏𝟎𝟎 𝑽𝑽 𝑽𝑽 𝑽𝑽 𝑺𝑺𝟏𝟏 𝑹𝑹𝑻𝑻 𝑺𝑺𝟐𝟐 𝑹𝑹𝑻𝑻 𝑺𝑺𝟑𝟑 𝑹𝑹𝑻𝑻 𝑷𝑷𝟏𝟏 = , 𝑷𝑷𝟐𝟐 = and 𝑷𝑷𝟑𝟑 = are the partial pressure of gases 1, 2 and 3 𝑽𝑽 𝑽𝑽 𝑽𝑽 respectively. Therefore,𝑷𝑷 = 𝑷𝑷𝟏𝟏 + 𝑷𝑷𝟐𝟐 + 𝑷𝑷𝟑𝟑 𝟔𝟔. 𝟏𝟏𝟏𝟏 In other words the total pressure exerted by a mixture of chemically non-reactive gases is equal to the sum of the partial pressures of the individual gases. This is Dalton’s law of partial pressure. 6.3 Solved Problems 48 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 1 1. A fixed mass of gas is 2000cm2. If the pressure is reduced to of its original value 5 while the temperature remains constant, what is its new volume? Solution: 𝑽𝑽𝟏𝟏 = 2000 𝑐𝑐𝑐𝑐3 , 𝑷𝑷𝟏𝟏 = 1 𝑋𝑋105 𝑁𝑁/𝑐𝑐2 , 1 𝑷𝑷𝟐𝟐 = (1 𝑋𝑋105 )𝑁𝑁/𝑐𝑐2 = 2 𝑋𝑋104 𝑁𝑁/𝑐𝑐2 5 𝑽𝑽𝟐𝟐 = ? 𝑽𝑽𝟏𝟏 𝑷𝑷𝟏𝟏 2000 𝑋𝑋 1 𝑋𝑋105 𝑽𝑽𝟐𝟐 = = = 𝟏𝟏𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 𝒔𝒔𝒎𝒎𝟑𝟑 𝑷𝑷𝟐𝟐 2 𝑋𝑋104 2. The gauge pressure at S.T.P. of a certain amount of gas occupying a volume of 0.05 𝑐𝑐3 at a temperature of 270C is 1.99 𝑋𝑋 105 𝑃𝑃𝐷𝐷. Calculate the new gauge pressure if the volume is decreased to 0.02 𝑐𝑐3 and the temperature increased to 127 0C Solution: 𝑉𝑉1 = 0.05 𝑐𝑐3 𝑇𝑇1 = 27 + 273 = 300 𝐾𝐾 𝑃𝑃1 = 9.8 𝑋𝑋 104 + 1.01 𝑋𝑋 105 = 1.99 𝑋𝑋 105 𝑃𝑃𝐷𝐷 𝑉𝑉2 = 0.02 𝑐𝑐3 𝑇𝑇1 = 127 + 273 = 400 𝐾𝐾 𝑃𝑃2 = ? 𝑷𝑷𝟏𝟏 𝑽𝑽𝟏𝟏 𝑷𝑷𝟐𝟐 𝑽𝑽𝟐𝟐 = 𝑻𝑻𝟏𝟏 𝑻𝑻𝟐𝟐 𝑷𝑷𝟏𝟏 𝑽𝑽𝟏𝟏 𝑻𝑻𝟐𝟐 1.99 𝑋𝑋 105 𝑋𝑋 0.05 𝑋𝑋 400 𝑷𝑷𝟐𝟐 = = = 𝟔𝟔. 𝟔𝟔𝟑𝟑 𝑿𝑿 𝟏𝟏𝟎𝟎𝟓𝟓 𝑷𝑷𝑺𝑺 𝑻𝑻𝟏𝟏 𝑽𝑽𝟐𝟐 300 𝑋𝑋 0.02 49 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 6.4 Supplementary Problems 1. Gas in a tank has a gauge pressure of 2.202 𝑋𝑋 105 𝑃𝑃𝐷𝐷 and volume 4 𝑐𝑐3 at 27 0C. Calculate the gauge pressure when the gas is compressed to 0.025 𝑐𝑐3 and the temperature has risen to 40 0C. 2. Gas occupying a container has a pressure of 1.5 𝐷𝐷𝐷𝐷𝑐𝑐𝑠𝑠 at 450C. Calculate the gauge pressure when the container and its content are cooled to 00C. Assume the change in volume of container is negligible. 3. A certain mass of hydrogen gas occupies 370 𝐿𝐿at 16 0C and 150 𝑃𝑃𝐷𝐷Find its volume at −21 0C and 420 𝑃𝑃𝐷𝐷 6.5 The Kinetic Theory of Gases The molecular theory of matter is based on the premise that all matter is made up of tiny particles called molecules. The molecules of a solid are tightly packed with little spaces between them and they are held together by forces of attraction (cohesive forces). In the solid state the molecules are not free to move around; they only vibrate about their mean positions. The molecules of liquid are more loosely-packed than those of a solid and the cohesive forces are sufficiently weak to enable the liquid molecules to slide past one another although they still remain quite close together. The molecules of a gas are much more loosely-packed than those of a liquid and they move about randomly at very high speeds (more than 1500 Km/hrs. for molecules of air at normal temperature and pressure). When confined in a container these molecules frequently collide with one another and with the walls of the container. The rapid and continuous collisions of the fast- moving molecules with the wall of the container give rise to the effect know as pressure. The behaviour of gases as revealed by the gas laws can be understood by considering a gas which is confined in a cubical container of length L (Figure 6.3). The following assumptions are made in the kinetic theory: 50 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 1. The gas is composed of a large number of identical molecules all moving about randomly and colliding with one another and wit the walls of the container 2. The molecules are so small that the volume of individual molecules could be neglected, i.e. the molecules can be regarded as point masses each of mass m and negligible volume. 3. The collisions of molecules with each other and with the walls of the container are perfectly elastic i.e. the total kinetic energy of the molecules before collision is equal to the kinetic energy after collision. 4. The force of attraction (or repulsion) between molecules is negligible except during collisions. Figure 6.3: Gas Molecule confined cubical container Any of the gas molecule facing the container experiences a pressure due to the bombarding gas molecules which is equal to the total force exerted on that face divided by its area, L2. Consider a molecule of mass m (Fig ) which moves with velocity 𝒗𝒗having components 𝒗𝒗𝒙𝒙 , 𝒗𝒗𝒅𝒅 and 𝒗𝒗𝒛𝒛 in the x, y and z directions respectively. The molecule moves towards the face EFGH with an x-component of velocity𝒗𝒗𝒙𝒙. After colliding with the wall EFGH, the molecule starts moving in the opposite direction (with velocity –𝒗𝒗𝒙𝒙 ) towards the opposite direction face ABCD. Upon hitting ABCD it again starts moving with velocity 𝒗𝒗𝒙𝒙 toward EFGH, and so on. 51 Downloaded by Donald Ngorka ([email protected])  lOMoARcPSD|33130136 The force exerted on EFGH by any one molecule is equal to the average rate of change of momentum of the molecule, or the change of momentum during each collision divided by the time between collisions. The change in momentum of a molecule whose velocity is changed from 𝒗𝒗𝒙𝒙 to -𝒗𝒗𝒙𝒙 as a result of collision is equal to m𝒗𝒗𝒙𝒙 – (-m𝒗𝒗𝒙𝒙 ) or 2m𝒗𝒗𝒙𝒙. The time interval (t) between two collisions is the distance traveled by the molecule between the two collisions divided by its velocity, i.e. 𝟐𝟐𝒍𝒍 𝑺𝑺 = 𝟔𝟔. 𝟏𝟏𝟐𝟐 𝒗𝒗𝒙𝒙 (Note that a molecule after colliding with wall EFGH, travels to wall ABCD and back before the next collision, covering distance 𝟐𝟐𝒍𝒍 in the process) 𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜 𝐢𝐢𝐜𝐜 𝟒𝟒𝐜𝐜𝟒𝟒𝐜𝐜𝟒𝟒𝐦𝐦𝐦𝐦𝟒𝟒 𝟐𝟐𝒎𝒎𝒗𝒗𝒙𝒙 𝒎𝒎𝒗𝒗𝟐𝟐𝒙𝒙 Average force on EFGH = = = 𝟔𝟔. 𝟏𝟏𝟑𝟑 𝐦𝐦𝐢𝐢𝟒𝟒𝐜𝐜 𝐛𝐛𝐜𝐜𝐦𝐦𝐛𝐛𝐜𝐜𝐜𝐜𝐜𝐜 𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐢𝐢𝐜𝐜𝐢𝐢𝐜𝐜𝐜𝐜𝐜𝐜 𝒍𝒍/𝒗𝒗𝒙𝒙 𝒍𝒍 For N molecules each moving with different velocities, the total force (F) on EFGH is given by 𝑵𝑵𝒎𝒎𝒗𝒗 𝟐𝟐 𝒙𝒙 𝑭𝑭 =