4th Class Chapter 4 Thermal Properties of Materials PDF

4th Class Chapter 4: Thermal properties of materials Physics Quantity of Heat and Specific Heat D Q: What is the internal energy heat of matter? An: it is the sum of the kinetic and the...

4th Class Chapter 4: Thermal properties of materials Physics Quantity of Heat and Specific Heat D Q: What is the internal energy heat of matter? An: it is the sum of the kinetic and the potential energy of the particles of the matter. Q: What is the quantity of heat? What its unit? On what does it depend? An: It is the amount of heat needed to heat an object. It is measured in Joule (J) or in calorie unit and one calorie equals to (4.2J). It depends on; 1.Mass of the object 2. The change in its temperature 3. The type of the material it is made of Q: What is the ⑮heat? heat? specific An: It is the quantity of heat required to raise the temperature of (1kg) from material one Celsius degree and its unit is (Joule/kg.°C). It symbol is (CP). Note: 1- We can calculate the quantity of heat (Q) needed to raise the temperature of an object with mass (m) from a certain temperature (T1) to a temperature (T2) by the following relation; Heat quantity = Mass of the object × Material Specific heat × Change in temperature 2- The sign of (ΔT) and (Q) is positive when the material gains heat energy from the surrounding so its temperature increases and the sign is negative when the material losses heat energy to the surrounding so its temperature decreases. Q: What is heat capacity? On what does it depend? An: The amount of heat required raising the temperature of the whole mass from material one degree Celsius and its unit is (Joule/°C). Its symbol is (C). Heat Capacity = Mass of the Object × Specific Heat ⑮ It depends only on the type of material, object mass and the specific heat for its material. Q: What is the relation between heat quantity and specific heat? An: I Since then That means; Heat Quantity = Heat Capacity × Change in Temperature 1 4th Class Chapter 4: Thermal properties of materials Physics Example 1 What is the amount of heat energy required to raise the temperature of (3kg) aluminum from (15°C) to (25°C) noting that the specific heat of aluminum is (900 J/kg.°C). Solution: Mass of aluminum = 3 kg Initial temperature (before heating) for the aluminum T1= 15 °C. Final temperature (after heating) for the aluminum T2 = 25 °C. Specific heat of aluminum Cp = 900 (J/kg.°C). And according to the equation; Q = m CP (T2-T1). Q = 3kg × 900 J/kg.°C× (25-15)°C. Q = 27000 J amount of heat energy. Q: What is the benefit of being the specific heat for water is greater than all other material used in our daily life? An: This can help us to explain many natural phenomenon. And it is also useful in much daily application as: 1. Its Effect to the weather (land and sea breeze). 2. Its use in car engine cooling. 3. Cooling of the machines in the factories using water. Example 2 What is the heat capacity for a piece of iron of (4kg) mass and its specific heat is (448 J/kg.°C)? Solution: Heat capacity = mass × specific heat C = m CP C = 4kg × 448 J/kg.°C = 1792 J/°C the heat capacity Question If you have three different metal pieces and where given the same heat quantity so their tem- perature increased as shown in the following figure then which one of them have the greater heat capacity, explain your answer? An: the metal of ΔT = 3oC has the greater heat capacity because heat capacity is in inverse proportion with ΔT 2 4th Class Chapter 4: Thermal properties of materials Physics Thermal Equilibrium Q: What is it meant by thermal equilibrium? An: when mixing two liquids together heat moves from the hot object to the cold one and the heat flow continues until the temperature of the two liquids become equal and results in a thermal equilibrium in the isolated system. Heat Lost = Heat Gained Q: What is the calorimeter? What does it compose of? An: It is as a water container that is thermally insulated used to measure the specific heat for a certain object. it is composed of a thin container made up of a good heat conducting metal like copper and its surrounded by another container from the same metal and is separated by a heat insulating material like sawdust or fibers to thermally insulate the inner container and its contents from its surrounding medium and it has a cover, with two aperture first to enter the thermometer and second to enter the stirrer to mix the material together Example 3 An aluminum cube of (0.5) kg at (100°C) was placed in a container filled with (1kg) of water at (20 °C), (suppose that there is no lose for the heat energy to the surrounding), calculate the final temperature for the (aluminum and water) when thermal equilibrium is reached (means the temperature of the aluminum and water equalize). (Noting that the specific heat for water is (4200 J/kg.°C) and specific heat of aluminum is (900 J/kg.°C) Solution Suppose the final temperature for them = Tf °C. So the aluminum temperature decreases by (100- Tf ) °C. And the water temperature increases by (Tf - 20 ) °C. We apply the following equation: Heat Lost = Heat Gained Water = W, Aluminum = A mw × Cpw (Tf - 20)w = mA × CpA (100 - Tf )A 1 × 4200 (Tf - 20) = 0.5 × 900 × (100 - Tf ). 4200 Tf – 84000 = 45000 - 450 Tf Tf = 129000 / 4650 Tf = 27.7 °C the final temperature for both of them. 3 4th Class Chapter 4: Thermal properties of materials Physics Example 4 Calculate the heat capacity for a copper calorimeter contains (100g) of water at a temperature (10°C) add another amount of water its mass of (100g) at a temperature (80°C) and the mixtures final temperature became (38°C)? Solution We assume that the heat capacity for the calorimeter is C Quantity of gained heat Quantity of heat the cold water gained = mass × specific heat of water × change in temperature Q1= m Cp (T2 - T1) =0.1× 4200 × (38-10) =11760 J Amount of heat the water gained Quantity of heat the calorimeter gained = heat capacity of the calorimeter × change in temperature Q2 = C (T2 - T1) Q2 = C(38-10) = 28 C Quantity of heat the hot water lost = mass × specific heat × change in temperature Q3 = m Cp × (Tf - T1) = 0.1× 4200 × (38-80) = - 17640J During thermal equilibrium Quantity of gained heat (Q2 + Q1) = Quantity of lost heat (Q3) Quantity of heat that the water and the calorimeter gained = quantity of heat that the hot water lost Q3 = Q1 + Q2 17640 = 11760 + 28C C = 5880/28 = 210 J/°C heat capacity of the calorimeter Heat Effect on Materials Q: How does heat affect the materials? An: When the temperature of the solid, liquid or gas material is increases the average kinetic energy of the particles increases so the gaps between them increase then it gets in expansion Q: What is it meant by heat expansion? An: Expansion means increase in the dimensions of the material Q: on what does the heat expansion depend? Explain An: It depends on the state of the material, gases expansion is more than liquids and liquids expansion is more than solids if the gained heat was equal for the three states. Expansion of solids; It is occurs in three forms; 1- Longitudinal expansion means increase in the length (expansion in one dimension) 2- Surface expansion means increase in the surface area (expansion in two-dimensions). 3- Volumetric expansion means expansion in objects volume (expansion in three-dimensions) 4 4th Class Chapter 4: Thermal properties of materials Physics Q: On what does the thermal change in length depend? An: the change in the length is directly proportional with the change in temperatures and original length and type of material. The equation of the length change can be written as the following: Where; ΔL is the change in length, is longitudinal expansion factor, L original length and ΔT is the change in temperature Q: What is the longitudinal expansion factor? An: The amount of increase in the unit lengths of the material when heated one degree Celsius is measured by ( ) unit and it differs according to the material; Q: Write the equation of change the area by heat? Give the physic meaning of each symbol An: Where; ΔA is the change in area, is surface expansion factor, A original area and ΔT is the change in temperature Q: Define the surface expansion factor? An: The amount of increase in the area unit from object when temperature rises one degree Celsius and is measured by (1/°C) unit. Q: How does volume of materials change by change the temperature? An: The change in the volume of the matter as the temperature changes and it’s described by volumetric expansion factor of matter (β), Thus the volume of the matter (V) increases by (ΔV) amount due to temperature increase by (ΔT) amount; Where; ΔV is the change in volume, is volumetric expansion factor, V original volume and ΔT is the change in temperature Q: What is volumetric expansion factor? An: It is amount of increase in the volume unit of the material when temperature rises one degree Celsius and is measured by (1/°C) unit. Q: What is the relation between longitudinal expansion factor and the surface expansion factor and volumetric expansion factor? An: 5 4th Class Chapter 4: Thermal properties of materials Physics Applications of solid material expansion by heat:- Q: State some applications of solid material expansion by heat? An: 1- Automatic thermostat in electrical devices like fridge, freezer, flat irons and fire alarm device. A bimetallic strip band is used to control the opening and the closing of the electric circuit. 2- In the electric lamp industry. The lamp’s glass and the wire have equal thermal expansion factor. So that they expand by the same amount in order to prevent the lamp’s base from breaking. 3- In the construction design to avoid the dangers and that’s by putting proper spaces or sepa- rators in bridges and leaving spaces between railway bars. Thermal expansion of liquids Q: What is the effect of heat on liquids? An: Liquids expand by increasing the temperature. Because the heat increases the average kinetic energy of the particles and increases the gaps between them then it gets the liquid expanded. Q: Why does the liquid in narrow glass tube drop then rise when it is placed in a container of hot water? An: the level of the water drops a little in the tube because the Aclass g expands at first and increases in the volume, so the water level drops. When the heat reaches the water it expands and raises in the tube due to its volume increase. Noting that; the volumetric expansion of liquids is greater than the volumetric expansion of solid materials under the same temperature change. Q: What is it meant by; real expansion and apparent expansion of liquid? An: real expansion is the expansion of liquid under the effect of heat (expansion of liquid without container). Apparent expansion is the expansion of liquid and the container that contains the liquid. Real expansion of liquid is always larger than apparent expansion. Q: Why does the Pyrex glass bear fast changes in temperature without breaking? An: Because longitudinal expansion factor of Pyrex glass is smaller compared to normal glass. Example The gasoline tank of the car of (60) litter volume was fully filled with gasoline when the temperature was (25°C), then the car was left under the sun light many hours till the tank temperature became (45°C), calculate the volume of the gasoline that’s expected to spill from the tank (neglect the gasoline expansion)? Expansion factor for gasoline is: β = 9.6 × 10-4 1/°C Solution ΔT = T2 - T1 = 45 - 25 = 20 °C ΔV = V β ΔT ΔV = 60 × 9.6 × 10-4 × 20 ΔV = 1.152 Litter the volume of the spilled gasoline. 6 4th Class Chapter 4: Thermal properties of materials Physics Gases expansion Q: Why gases expansion is greater than the liquids expansion and the solid expansion? An: because gases have molecular force between their molecules less than the force between molecules of solid and liquid. Q: what is gases expansion characterized? An: gases are characterized by having equal volumetric expansion factor at constant pressure. Q: Why is the apparent expansion of gases considered as real expansion? An: Because the expansion of the container that contain the gas is very small compared to the expansion of the gas itself so container’s expansion can be neglected thus the apparent expansion of gases is considered as real expansion. State Change of Matter Q: What is it meant by; melting point, freezing point and boiling point? An: Melting point it is the temperature which matter start converting from solid state to liquid state. Freezing point it is the temperature which matter start converting from liquid state to solid state. Boiling point it is the temperature which matter start converting from liquid state to gas state. Note: 1- Melting (fusion) point = Freezing point 2- Melting point of water = 0°C = Freezing point of water 3- Boiling point of water = 100°C Latent heat of fusion: the quantity of heat required to convert a masses unit from solid to liquid at the same temperature (for example melting point for ice is 0°C) and under constant pressure is called latent heat of fusion and measured by (J/kg) kJ/kg units. B Quantity of heat required to melt the material = mass × latent heat of fusion Where; m represents mass of the object, Lf represents latent heat of fusion. Example Calculate the quantity of heat required to convert an ice piece of (25g) at (0°C) to water under the same temperature. Solution Amount of heat = mass × latent heat of fusion. Q = m Lf = (25/1000) × 335 Q = 8.375 kJ the quantity of heat required 7 4th Class Chapter 4: Thermal properties of materials Physics Example Calculate the quantity of heat required to convert (2 kg) of ice at (-15 °C) to water at (25 °C) noting that specific heat of water is (4200 J/kg.°C) and the latent heat of fusion for ice at (0 °C) is (335 kJ/kg) and specific heat of ice is (2093 J/kg.°C). Solution To raise the ice temperature from (-15 to 0 °C) we need a heat quantity equals to: Quantity of heat = mass × specific heat of ice × temperature difference Q1 = m Cice ΔT = 2 × 2093 × [0-(-15)] = 2 × 2093 × 15 = 30 × 2093 Q1 = 62790 Joule To convert ice to water at (0 °C) we need a heat quantity equals to: Quantity of heat = mass × latent heat of fusion of ice Q2 = m Lf = 2× 335 kJ/kg Q2 = 670000 Joule To raise the water temperature from (0 to 25 °C) we a heat quantity equals to: Quantity of heat = mass × specific heat of water × temperature difference Q3 = m Cwater × ΔT = 2 × 4200 × (25-0) = 50 × 4200 Q3 = 210000 Joule To calculate the heat quantities that were supplied to the ice till it became water equal (25 °C): Qtotal = Q1+ Q2 + Q3 = 62790 + 670000 + 210000 Qtotal = 942790 Joule total heat quantity Latent heat of vaporization; The heat quantity required to convert a masses unit of the material from liquid to gas state at the boiling point is called latent heat of vaporization. A v Where; m represents mass of the object, Lv represent the latent heat of vaporization and measured by kJ/kg. Example Calculate the quantity of heat required to convert (3kg) of water at (20°C) to vapor at (110°C) noting that the specific heat of water is (4200 J/kg) and the latent heat of vaporization of water is (2260 kJ/kg) and the specific heat of water vapor is (2010 J/kg°C) ? Solution Total heat quantity = Quantity of heat required to heat the water from (20 to 100°C) + quantity of heat required to convert water to vapor at 100°C + quantity of heat required to raise the water vapor temperature from (100 to 110°C). Qtotal = Q1 + Q2 + Q3 = mc (T2-T1) + mLv + mc (T3-T2) = 3 × 4200 × (100-20) + 3 × 2260 × 103 + 3 × 2010 × (110-100) = 1008000 + 6780000 + 60300 Qtotal = 7848300 J total heat quantity 8  4th Class Chapter 4: Thermal properties of materials Physics Methods of Heat Transfers Q: How does heat transfer? An: heat transfers from one object to another by three ways 1. Conduction 2. Convection 3. Radiation Q: Explain how does heat transfer in solid materials? An: heat transfers in solid material by conduction method, and the time average of the transferred thermal energy differs from one material to another depending on the internal structure of the material Q: Why are metals considered a good thermal conductor? An: metals are considered good thermal conduction materials and that is because of their free electrons and close atoms while heat transfers very weekly in poor thermal conduction materials like wood, rubber, plastic and others. Thermal conductivity Q: What is thermal conductivity? An: It is a property of transfer thermal heat through an object by conduction. Q: What is the thermal gradient? An: The amount of change in conductor’s temperature at each meter of its length when the heat transfers perpendicular on its cross sectional area L is the length of the object Q: How does thermal energy change with length of conductor? An: from the equation ( ⑱ ) we find that as long as the length thermal B gradient increases the amount of thermal energy flow ⑱ increases. (Inverse proportion) decreases. Q: State the equation of the time average for the transfer of thermal energy? An: Where; H: represents the time average for the transfer of thermal energy by conduction method and measured by (Watt) A: cross section area measured by (m2). ΔT : temperature differences measured by (°C). L: length of the rod (or its thickness) measured by (m). k: thermal conduction coefficient measured by (watt/m.°C). Note: different solid materials have different thermal conduction coefficients 9 4th Class Chapter 4: Thermal properties of materials Physics Example page 63 An iron rod of (50cm) length and sectional area of (1cm2), its one side was placed on flame at (200°C) and the other end was placed in crushed ice at (0°C), if the rod was covered by insulating material and if the thermal conduction coefficient of iron is (79 Watt/m.°C) then calculate: 1. Thermal gradient 2. The time average for the transfer of thermal energy. Solution 1- 2- Example page 64 Room has a glass window with one-layer, if the length of the window (2.2m), width (1.2m) and thickness 5mm, assuming the temperature of the window surface inside the room (22°C) and the temperature outside it (3°C), calculate the time average to transfer thermal energy from room, note the thermal conduction coefficient of glass (0.8w/m.°C)? Solution Thermal conduction applications 1. Using metals to make Kitchen-ware. 2. Using insulating material for handles in Kitchen-ware. 3. Thermal insulation in house constructions using insulating materials like air, glass and polystyrene 4. The thermos can Q: What is the thermos can? An: It is a container made of an interior layer of plastic and an exterior layer of polystyrene, and according to this system the temperature of the liquid in it is kept by reducing thermal leakage to the outside. Question If an ice cube was placed in an aluminum box and another identical cube was placed in a wood box, then which one of the cubes melts first at room temperature? An: Ice cube in aluminum box melts first because the thermal conduction coefficient of aluminum is greater than the thermal conduction coefficient of wood. Transfer of heat by convection Q: What is the difference between transfer heat by conducting and convection? An: in the thermal conduction method the thermal energy flows through the material without the transfer of the material’s particles themselves. It occurs in solid only But in thermal convection method that the material’s particles themselves move and transfer from one place to another, and thermal convection occurs only in fluid and doesn’t occur in solids 11 4th Class Chapter 4: Thermal properties of materials Physics Types of convection Q: What are the types of convection? An: 1- Free convection 2- Forced convection Q: Explain what is it meant by free convection? An: Free convection In this type the convectional current generate by effect of gravity cold air (or cold liquid) have greater density so it comes down because the up-thrust force is less than its weight, while hot air’s density is less so it goes up carrying the thermal energy with it because the up-thrust force is greater than its weight in this case. Forced convection in this type the fluid is motivated to circulate by installing a pump or a fan in the fluid’s way that creates a pressure difference which forces the particles to move, hence in some central heating processes the hot air is pushed into the halls using fan or the hot water is pumped into radiators that’s placed on the ground. Heat transfer by radiation: Heat is transferred by radiation in form of electromagnetic waves in the speed of light. Q: On what the wavelength of electromagnetic wave depend? An: It depends on the temperature of radiating object Q: On what is the amount of radiation energy emitted from the objects depend? An: 1. The nature of the surface that emits radiation energy that’s as long as the surface’s area increases the amount of emitted energy increases. also its color, since black surface radiates energy with an average that is very greater then the average emitted by a fair colored object. 2. Temperature: the objects radiates energy in the form of electromagnetic waves that can be seen if the temperature of the objects was high while the radiations are invisible when the object’s temperature is low. Note: Material those are good heat radiation are also good heat absorbers Q: Why is the amount of absorbed heat energy by materials different? An: the amount of absorbed heat energy differs by the following changes: 1. Type of the material 2. Color of the material 3. How smooth it is; where smooth and fair colored objects absorb radiation energy less than rough and dark colored objects do Application on heat transfer by convection and radiation methods: 1. Plastic (glass) houses; radiation, 2. Solar heater; radiation 3. Central heating; convection + radiation 4. Night photographing by infrared rays; radiation. Thermal Pollution: It is human activities that result together in the temperature increase of the land and atmosphere and water which leads to a dysfunction in the environmental structure. Thermal pollution sources: 1. Electrical energy generating sources 2. Petroleum industry and refineries 11