Heat and Temperature Lecture 1 (PDF)

 Heat Lecture 1 Heat and Temperature  Heat Heat is a form of energy that is transferred from one body to another as the result of a difference in temperature. Unit of heat energy : Joule  Temperature How hot or cold an object feels when...

 Heat Lecture 1 Heat and Temperature  Heat Heat is a form of energy that is transferred from one body to another as the result of a difference in temperature. Unit of heat energy : Joule  Temperature How hot or cold an object feels when we touch it.  When two objects are placed in contact heat (energy) is transferred from one to the other until they reach the same temperature (are in thermal equilibrium). When the objects are at the same temperature there is no heat transfer. TA < TB TA = TB (Thermal Equilibrium)  Zeroth Law of Thermodynamics If two objects A and B are separately in thermal equilibrium with a third object C, then A and B are in thermal equilibrium with C. 𝑨⇄𝑪 𝒂𝒏𝒅 𝑩⇄𝑪 ∴𝑨⇄𝑩  Thermometers and Temperature Scale Thermometers are devices that are used to measure the temperature of a system. All thermometers are based on the principle that some physical property of a system changes as the temperature changes. A common thermometer in everyday use consists of a mass of liquid usually mercury or alcohol that expands into a glass capillary ( ‫ )شعريه‬tube when heated. Some examples of physical properties that change with temperature are (1) the volume of a liquid, (2) the pressure of a gas at constant volume, (4) the volume of a gas at constant pressure, (5) the electric resistance of a conductor, and (6) the color of an object.  Types of the temperature scales a. Celsius scale Tc The boiling point of water = 100 oC The freezing point of water = 0oC a. Kalvin Scale TK 𝐓𝐊 = 𝐓𝐜 + 𝟐𝟕𝟑 c. Fahrenheit Scale TF 𝟗 𝐓𝐅 = 𝐓𝐜 + 𝟑𝟐 𝟓 The relationship between changes in temperature: 𝟓 ∆𝑻𝒄 = ∆𝑻 = ∆𝑻𝑭 𝟗 Ex: The boiling point of water on the Fahrenheit scale is…………. a) 32 oF b) 0 oF c) 100 oF d) 212 oF 𝟗 Ans : 𝐓𝐅 = (𝟏𝟎𝟎) + 𝟑𝟐 = 212oF 𝟓 Thermal expansion of solids and liquids When an object is heated or cooled, its dimensions change...‫االجسام تتمدد بالتسخين وتنكمش بالبروده‬ Types of Thermal expansion: a. Linear expansion: If the change in one dimension. b. Surface expansion: If the change in two dimensions. c. Volume expansion: If the change in three dimensions. a. Linear expansion: ‫التمدد الطولي‬ If we heat a rod of metal from T1 to T2: T1 original or initial length 𝒍𝟏 ∆𝑙 T2 Final length 𝒍𝟐 Change or increase in length : ∆ 𝒍 = 𝜶 𝒍 ∆𝑻 𝟏 𝛼 is the coefficient of linear expansion…..Unit: oC-1 𝒍𝟐 = 𝒍𝟏 + ∆ 𝒍 = 𝒍𝟏 + 𝜶 𝒍𝟏 ∆𝑻 ‫معامل التمد الطولي وهو ثابت يعتمد على طبيعه او‬ ‫نوع الماده ويمكن تعريفه علي انه الزياده في طول‬ The final length: 𝒍𝟐 = 𝒍𝟏 (𝟏 + 𝜶∆𝑻).‫الماده لكل زياده في درجه الحراره واحد درجه مئويه‬ b. Surface expansion: ‫التمدد السطحي‬ Consider a rectangular area of dimensions 𝑙1 and 𝑤1 and its temperature increases from (T1) To (T2): T2 T1 change or increase in area: 𝑙1 A1 ∆ 𝑨 = 𝝈 𝑨𝟏 ∆𝑻 A2 w1 𝜎: Coefficient of surface expansion; 𝝈 = 𝟐𝜶 𝒐𝒓 𝑨𝟐 = 𝑨𝟏 (𝟏 + 𝝈∆𝑻) c. Volume expansion: ‫التمدد الحجمي‬ Consider a volume has dimensions 𝒍𝟏 , 𝒘𝟏 and h1 its temperature increases from (T1) To (T2): h2 h1 Heat T1 T2 w1 w2 l1 l2 change or increase in volume: ∆ 𝑽 = 𝜷 𝑽𝟏 ∆𝑻 𝜷: Coefficient of volume expansion; 𝜷 = 𝟑𝜶 𝒐𝒓 𝑽𝟐 = 𝑽𝟏 (𝟏 + 𝜷∆𝑻)  Unusual expansion of water Most liquid expand when heated and contract when cooled, water, however is an exception. Between 0 ∘C and 4 ∘C, water expands when temperature decreases. Water has the highest density and smallest volume at 4 ∘C. The density become less and less as it freezes. This is why ice at 0∘C is less dense than water. Example: a. Convert -2.0 oC to degree Fahrenheit. b. Convert 59 oF to degree Celsius. Solution: 𝟗 𝟗 a. 𝑻𝑭 = 𝑻𝒄 + 𝟑𝟐 = −𝟐. 𝟎 + 𝟑𝟐 = 𝟐𝟖. 𝟒 o𝑭 𝟓 𝟓 𝟗 b. 𝑻𝑭 = 𝑻𝒄 + 𝟑𝟐 = 𝟓𝟗 o𝑭 𝟓 = (59-32)×5/9 = 15 o𝐂 Example: Aluminum has a thermal expansion coefficient of 25×10-6 oC-1. By how much would the length of a 1.0 meter bar change if it were subjected to a temperature change of +25 oC? Solution: ∆ 𝑙 = 𝛼 𝐿1 ∆𝑇 = 25 × 10-6 oC-1 × 1.0 m ×25 oC = 6.3 × 10 -4 m Example: A glass jar (α= 0.3x10-5/oC) has a metal lid (α = 1.6x10- 5/oC )which is stuck. If you heat them by placing them in hot water, the lid will be: a. Easier to open b. harder to open c. the same