Lecture 1 - Heat PDF

Chapter 1: Heat Chapter 1 Temperature & Heat Temperature (T): Temperature is a measure for the degree of hotness or coldness of the bodies. Temperature scales The temperature is measured in many scales, The mo...

Chapter 1: Heat Chapter 1 Temperature & Heat Temperature (T): Temperature is a measure for the degree of hotness or coldness of the bodies. Temperature scales The temperature is measured in many scales, The most common types are: 1- Celsius scale (0C) 2- Fahrenheit scale (0F) 3- Kelvin scale (K) Where: - The freezing point of water is: 0 0C = 32 0F = 273.15 K - The boiling point of water is: 100 0C = 212 0F = 373.15 K 1  Chapter 1: Heat TC is a certain temperature in the Celsius scale while TF and TK are the corresponding temperatures for TC in Fahrenheit and Kelvin scales. To find the relation between these scales, one can use the following relation: TC −0 F T −32 K T −273.15 ⸫ = = 100−0 212−32 373.15−273.15 𝐓𝐂 𝐓𝐅 −𝟑𝟐 𝐓𝐊 −𝟐𝟕𝟑.𝟏𝟓 ⸫ 𝟏𝟎𝟎 = 𝟏𝟖𝟎 = 𝟏𝟎𝟎 - Relation between TC and TF: T C TF −32 ⸪ 100 = 180 TF −32 ⸫ TC = 100 x 180 𝟓 ⸫ 𝐓𝐂 = (𝐓𝐅 − 𝟑𝟐) 𝟗 Or 𝟗 ⸫ 𝐓𝐅 = ( 𝐓𝐂 ) + 𝟑𝟐 𝟓 For two different temperatures in Celsius scale TC1 and TC2 (as: TC2 > TC1) and their corresponding temperatures in Fahrenheit scale TF1 and TF2 (as: TF2 > TF1) 5 TC1 = (T − 32) 9 F1 And 5 TC2 = (T − 32) 9 F2 2  Chapter 1: Heat ⸫ The change in temperatures in Celsius scale ΔTC, is: ΔTC = TC2 – TC1 5 ⸫ ΔTC = (TF2 − TF1 ) 9 𝟓 ⸫ 𝚫𝐓𝐂 = 𝚫𝐓𝐅 𝟗 Where, ΔTF = TF2 – TF1 ≡ corresponding change in temperature in Fahrenheit scale - Relation between TC and TK: T C 𝐓𝐊 −𝟐𝟕𝟑.𝟏𝟓 ⸪ 100 = 𝟏𝟎𝟎 ⸫ 𝐓𝐂 = 𝐓𝐊 − 𝟐𝟕𝟑. 𝟏𝟓 Or ⸫ 𝐓𝐊 = 𝐓𝐂 + 𝟐𝟕𝟑. 𝟏𝟓 For two different temperatures in Celsius scale TC1 and TC2 (as: TC2 > TC1) and their corresponding temperatures in Kelvin scale TK1 and TK2 (as: TK2 > TK1) TC1 = TK1 − 273.15 And TC2 = TK2 − 273.15 ⸫ The change in temperature in Celsius scale ΔTC, is: ΔTC = TC2 – TC1 ⸫ ΔTC = TK2 − TK1 ⸫ 𝚫𝐓𝐂 = 𝚫𝐓𝐊 Where, ΔTK = TK2 – TK1 ≡ corresponding change in temperature in Kelvin scale 3  Chapter 1: Heat - Relation between TF and TK: ⸪ TK = TC + 273.15 and 5 ⸪ TC = (TF − 32) 9 𝟓 ⸫ 𝐓𝐊 = [ (𝐓𝐅 − 𝟑𝟐)] + 𝟐𝟕𝟑. 𝟏𝟓 𝟗 Or 𝟗 ⸫ 𝐓𝐅 = [ (𝐓𝐊 − 𝟐𝟕𝟑. 𝟏𝟓)] + 𝟑𝟐 𝟓 For two different temperatures in Kelvin scale TK1 and TK2 (as: TK2 > TK1) and their corresponding temperatures in Fahrenheit scale TF1 and TF2 (as: TF2 > TF1) 5 ⸪ ΔTC = ΔTF 9 and ⸪ ΔTC = ΔTK 𝟓 ⸫ 𝚫𝐓𝐊 = 𝚫𝐓𝐅 𝟗 Where, ΔTK = TK2 – TK1 ≡ change in temperature in Kelvin scale ΔTF = TF2 – TF1 ≡ corresponding change in temperature in Fahrenheit scale 4  Chapter 1: Heat Example 1: Convert the following temperatures from the Fahrenheit scale to the Celsius one: a) 78 0F b) 300 0F Solution 𝟓 ⸪ 𝐓𝐂 = (𝐓𝐅 − 𝟑𝟐) 𝟗 a) TF = 78 0F ➔ TC = ?? 5 ⸫ TC = (78 − 32) 9 ⸫ TC = 25.56 0C b) TF = 300 0F ➔ TC = ?? 5 ⸫ TC = (300 − 32) 9 ⸫ TC = 148.89 0C Example 2: Convert the following temperatures from the Celsius scale to the Kelvin one: a) 20 0C b) -65 0C Solution ⸫ 𝐓𝐊 = 𝐓𝐂 + 𝟐𝟕𝟑. 𝟏𝟓 a) TC = 20 0C ➔ TK = ?? ⸫ TK = 20 + 273.15 ⸫ TK = 293.15 K 5  Chapter 1: Heat b) TC = -65 0C ➔ TK = ?? ⸫ TK = −65 + 273.15 ⸫ TK = 208.15 K Example 3: Convert the following temperatures from the Fahrenheit scale to the Kelvin one: a) 82 0F b) 272 0F Solution 𝟓 ⸫ 𝐓𝐊 = [ (𝐓𝐅 − 𝟑𝟐)] + 𝟐𝟕𝟑. 𝟏𝟓 𝟗 a) TF = 82 0F ➔ TK = ?? 5 ⸫ TK = [ (82 − 32)] + 273.15 9 ⸫ TK = 300.92 K b) TF = 272 0F ➔ TK = ?? 5 ⸫ TK = [ (272 − 32)] + 273.15 9 ⸫ TK = 406.48 K 6  Chapter 1: Heat Example 4: A block of material is heated from 30 0C to 90 0C. Calculate the change in the temperature in: a) Celsius scale b) Fahrenheit scale and c) Kelvin scale Solution TC1 = 30 0C, TC2 = 90 0C a) ΔTC = ?? ⸪ ΔTC = TC2 – TC1 ⸫ ΔTC = (90 – 30) 0C ⸫ ΔTC = 60 0C b) ΔTF = ?? 𝟓 ⸪ 𝚫𝐓𝐂 = 𝚫𝐓𝐅 𝟗 9 ⸫ ΔTF = ΔTC 5 9 ⸫ ΔTF = x 60 0F 5 ⸫ ΔTF = 108 0F c) ΔTK = ?? ⸪ 𝚫𝐓𝐊 = 𝚫𝐓𝐂 ⸫ ΔTK = 60 K 7  Chapter 1: Heat Example 5: At what temperature does Celsius and Fahrenheit scales coincide? Solution TC = TF = T = ?? 𝟓 ⸪ 𝐓𝐂 = (𝐓𝐅 − 𝟑𝟐) 𝟗 5 ⸫T = (T − 32) 9 9 ⸫ T = T − 32 5 9 ⸫ T = T − 32 5 4 ⸫ T = − 32 5 ⸫ T = - 40 0 ⸫ Celsius and Fahrenheit scales coincide at - 40 0 i.e., TC = - 40 0C in Celsius scale corresponds to TF = - 40 0F in Fahrenheit scale 8  Chapter 1: Heat Heat energy (Q): Heat is a form of energy which can be transferred from a hot substance to a colder one. Heat energy (Q) is measured in calorie (cal), Kilocalorie (Kcal) or Joule (J) As: 1 cal = 4.18 J, 1 Kcal = 103 cal & 1 Kcal = 4180 J Heating an object: If we have a substance of mass m and initial temperature Ti gains an amount of heat energy Q, its temperature will be raised to a final temperature TF. (i.e., TF > Ti) ⸫ ΔT = TF - Ti Cooling an object: If we have a substance of mass m and initial temperature Ti loses an amount of heat energy Q, its temperature will be lowered to a final temperature TF. (i.e., TF < Ti) ⸫ ΔT = Ti – TF 9  Chapter 1: Heat - The amount of heat Q gained (or lost) by the substance, without changing its state (phase), is given as: 𝐐 = 𝐦 𝐒 𝚫𝐓 Where, m = mass of the substance (g or Kg) ΔT = The change in temperature (0C) cal Kcal J S = Specific heat of substance ( , and ) g.0 C Kg.0 C Kg.0 C Note: In the previous law, we will deal (in this course), with the Celius scale only. Specific heat of substance (S): 𝐐 S= 𝐦 𝚫𝐓 It is defined as: “The amount of heat gained (or lost) by the substance to change the temperature of a unit mass (1 g or 1 Kg) by 10C” - We must know that: The specific heat of water (Sw); is cal Kcal J Sw = 1 =1 = 4180 g.0 C Kg.0 C Kg.0 C Definition of calorie: “It is the amount of heat gained (or lost) by 1 g of water to change its temperature by 1 0C” 10  Chapter 1: Heat VERY IMPORTANT NOTE: We use the previous law [𝐐 = 𝐦 𝐒 𝚫𝐓], only if the substance maintains the same phase (state). Phase (or state) means: Solid, liquid or gas. Example 6: Calculate the amount of heat required to raise the temperature of 4 Kg of water from 20 0C to the boiling point. (Given that: The specific heat of water is 4180 J/Kg.0C) Solution m = 4 Kg, Ti = 20 0C, TF = 100 0C, Sw = 4180 J/Kg.0C, Q = ?? ⸪ 𝐐 = 𝐦 𝐒𝐖 𝚫𝐓 ⸫ Q = m SW (TF − Ti ) ⸫ Q = 4 x 4180 x (100 – 20) J ⸫ Q = 1.34 x 106 J Example 7: If 6.7 KJ of heat energy is supplied to 100 g of glass at initial temperature of 20 0C. Find the final temperature. (Given that: The specific heat of glass is 670 J/Kg.0C) Solution Q = 6.7 KJ ➔ Q = 6.7 x 103 J m = 100 g ➔ m = 0.1 Kg Ti = 20 0C 11  Chapter 1: Heat Sglass = 670 J/Kg.0C TF = ?? ⸪ 𝐐 = 𝐦 𝐒𝐠𝐥𝐚𝐬𝐬 𝚫𝐓 ⸫ 6.7 x 103 = 0.1 x 670 x ΔT ⸫ ΔT = 100 0C ⸪ ΔT = TF - Ti ⸫ 100 = TF – 20 ⸫ TF = 120 0C 12