Chapter 4&5: Temperature and Heat Transfer PDF

Chapter 4&5: Temperature and Heat & Heat Transfer and Internal Energy SuperviSor: Dr. Hadeer El-Hawary New ASSiut TechNoLogicAL UNiverSity December 14, 2024 Temperature and Heat  Summary: Temperature: A measure of the average kinetic energy of the molecules of a substance. Units: 𝐾𝑒𝑙𝑣𝑖...

Chapter 4&5: Temperature and Heat & Heat Transfer and Internal Energy SuperviSor: Dr. Hadeer El-Hawary New ASSiut TechNoLogicAL UNiverSity December 14, 2024 Temperature and Heat  Summary: Temperature: A measure of the average kinetic energy of the molecules of a substance. Units: 𝐾𝑒𝑙𝑣𝑖𝑛 (𝐾) – 𝐶𝑒𝑙𝑠𝑖𝑢𝑠 (°𝐶) – 𝐹𝑎ℎ𝑟𝑒𝑛ℎ𝑒𝑖𝑡 (°𝐹). Conversions: 9 Celsius to Fahrenheit: 𝑇 ℉ = × 𝑇(℃) + 32 5 5 Fahrenheit to Celsius: 𝑇 ℃ = × (𝑇(℉) − 32) 9 Celsius to Kelvin: 𝑇(𝐾) = 𝑇(℃) + 273 Heat: The amount of energy transferred from one body to another due to a difference in temperature between them. Units: Calorie (𝐶𝑎𝑙) or Joule (𝐽). Zeroth Law of Thermodynamics: If the systems are in thermal equilibrium, no heat flow will take place. 1 8  Summary: Coefficient of Linear Expansion: ∆L 𝛼= 𝐿 ∆T Coefficient of Volume Expansion: ∆V β= V ∆T Note: For a solid, the average coefficient of volume expansion is three times the average linear expansion coefficient: 𝛽 = 3𝛼. 𝑚 Number of moles 𝑛: 𝑛 = 𝑀 m: mass; M: molar mass of the substance (molar mass of each chemical element is the atomic mass expressed in 𝑔/𝑚𝑜𝑙). 1 8  Summary: The Ideal Gas Law: 𝑁 𝑃𝑉 = 𝑛𝑅T = 𝑅T= 𝑁𝑘𝐵 T 𝑁𝐴 𝑃 : Pressure in 𝑃𝑎 𝑉 : Volume in 𝑚3 𝑛 : Number of moles of gas (dimensionless) 𝑅 : 𝑅 = 8.314 𝐽/𝑚𝑜𝑙 ∙ 𝐾 𝑇 : Temperature in 𝐾 𝑁𝐴 : Avogadro’s number(= 6.02 × 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒/𝑚𝑜𝑙) 𝑁 : Total number of molecules 𝑘𝐵 : Boltzmann’s constant(= 1.38 × 10−23 𝐽/K) 1 8  Problem (1): A pot of water boils from 38°𝐶 to 100°𝐶. What is the change in temperature ( ∆𝑇 ) on the Kelvin scale? 1 8 Solution: We have 𝑇𝑖 𝐾 = 𝑇𝑖 ℃ + 273 = 38 + 273 = 311 𝐾 and 𝑇𝑓 𝐾 = 𝑇𝑓 ℃ + 273 = 100 + 273 = 373 𝐾 Thus, ∆𝑇 𝐾 = 𝑇𝑓 𝐾 − 𝑇𝑖 𝐾 = 373 − 311 = 62 𝐾  Problem (2): A cube of steel has a volume of 50 𝑚3 when the temperature is 20.0°𝐶. What is its volume when the temperature is 50.0°𝐶? 1 8 Solution: ∆V We have β = and we know 𝛽 = 11 × 10−6 ℃−1 V ∆T Now, β V ∆T = ∆V ⇒ ∆𝑉 = 11 × 10−6 × 50 × 50 − 20 = 0.0165 m3 Finally, we get ∆𝑉 = 𝑉𝑓 − 𝑉𝑖 ⇒ 𝑣𝑓 = ∆𝑉 + 𝑉𝑖 = 50 + 0.0165 = 50.0165 m3.  Problem (3): An ideal gas occupies a volume of 100 𝑚3 at 10°𝐶 and 100 𝑃𝑎. Find the total number of molecules of gas in the container. (𝑘𝐵 = 1.38 × 10−23 𝐽/K) 1 8 Solution: Since 𝑃𝑉 = 𝑁𝑘𝐵 T, we can write down the following calculation: 𝑃𝑉 100×100 𝑁= = ≈ 7.25 × 1025 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠. 𝑘𝐵 𝑇 10×1.38×10−23 Heat Transfer and Internal Energy  Summary: Specific Heat Capacity 𝒄 : the quantity of heat required to change the temperature of 1 gram of a substance by 1°C. 𝑄 = 𝑚 𝑐 ∆𝑇 m : Mass of substance in (𝑘𝑔) Q : The heat energy in (𝐽) ∆𝑇: The change in temperature (°𝐶) Unit of 𝑐 is 𝐽/𝑘𝑔°𝐶. Conversion: 1 𝐶𝑎𝑙𝑜𝑟𝑖𝑒 = 4.187 𝑗𝑜𝑢𝑙𝑒𝑠 Notes: The state or phase of matter (Solid, Liquid or Gas) is determined by its temperature. Heat transfer always occurs whenever a substance changes phase; by absorbing or releasing energy. 1 8  Summary: Latent Heat: The hidden energy that is supplied or extracted to change the phase of 1 𝑔 of a substance without changing its temperature or pressure.. 𝑄 = ±𝑚L 𝑚 : Mass of substance in (𝑘𝑔) 𝑄 : The heat energy in (𝐽) 𝐿 : The latent heat (𝐽/𝑘𝑔) Notes: Latent heat of fusion (𝐿𝑓 ) ⇒ Melting Latent heat of vaporization (𝐿𝑣 ) ⇒ Boiling 1 8  Summary: Energy Transfer Mechanisms: Conduction (By Touch) ∆𝑄 ∆𝑇 𝑃= = 𝑘𝐴 ∆𝑡 𝑑 P: Heat conduction (𝑊 = 𝐽/𝑠) Q: Heat energy (𝐽) t : Time (𝑠) A: Cross-sectional area (𝑚2 ) 𝑑: Length (𝑚) k: Thermal conductivity (𝑊/𝑚°𝐶) Convection (By Air or Liquid) Radiation (By Electromagnetic Waves) 1 8  Summary: Convection (By Air or Liquid): Energy transferred by the movement of a warm substance is said to have been transferred by convection Radiation (By Electromagnetic Waves): Stephan-Boltzmann law: P = 𝑒𝜎𝐴𝑇 4 P: Power radiation (𝑊) 𝜎 : Stephen-Boltzmann constant ( = 5.67 −8 −2 −4 × 10 𝑊𝑚 𝐾 ) A: Surface area (𝑚2 ) 𝑒: Emissivity (no units) T: Temperature in (𝐾). 1 8  Summary: First Law of Thermodynamics: Whenever heat flows in or out of a system, the gain or loss of thermal energy equals the amount of heat transferred, i.e., energy is conserved. ∆𝑈 = 𝑄 − W or 𝑑𝑈 = 𝛿𝑄 − 𝑃 𝑑V ∆𝑈: Change in internal energy (𝐽) 𝑄 ∶ Amount of heat energy (𝐽) 𝑊 ∶Work done on the system by the surroundings (𝐽) 𝑃 : External Pressure (𝑃𝑎) 𝑑𝑉: (infinitesimal) Volume element (𝑚3 ) 1 8  Summary: Types of Thermodynamic Systems: Open System: System exchanges both matter and energy with surroundings. Closed System: System exchanges only energy (No Matter) with surroundings. Isolated System: System neither its energy nor its matter can be exchanged with surroundings. 1 8  Problem (1): A 0.05 𝑘𝑔 some metal is heated to 200°𝐶 and then dropped into a calorimeter containing 0.4 𝑘𝑔 of water initially at 20°𝐶. The final equilibrium temperature of the mixed system is 22.4°𝐶. Find the specific heat of the metal (𝑐𝑤 = 4186 𝐽/𝑘𝑔℃). 1 8 Solution: In equilibrium state, we have conceptually 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 = 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 i.e., decreasing in the amount heat of metal = increasing in the amount heat of water The metal losses heat energy which absorbed by water. That is, water gets more heat energy Symbolically, −𝑄𝑚𝑒𝑡𝑎𝑙 = +𝑄𝑤𝑎𝑡𝑒𝑟 −𝑐𝑚 𝑚𝑚 ∆𝑇𝑚 = 𝑐𝑤 𝑚𝑤 ∆𝑇𝑤 The ENd!