Phys 1011-Chapter 9 Heat and Thermodynamics PDF

Chapter 9 Heat and Thermodynamics  The Concept of Temperature and Heat What is temperature? Qualitatively, based on our senses, TEMPERATURE is a physical quantity that tells us how warm or cold an object is. This qualitative definition does not answer the question “how hot is...

Chapter 9 Heat and Thermodynamics  The Concept of Temperature and Heat What is temperature? Qualitatively, based on our senses, TEMPERATURE is a physical quantity that tells us how warm or cold an object is. This qualitative definition does not answer the question “how hot is hot?” Also, our senses are misleading  The Concept of Temperature and Heat What is temperature? We need an operational definition of temperature. Consider two bodies A and B, A being hotter than B, thermally isolated from each other. What happens when thermal contact is established between A and B? The Concept of Temperature and Heat Heat is energy in transfer from a hot to a cold body. Objects are in thermal contact if heat can flow between them. Eventually, the energy flow stops. The two bodies will be in thermal equilibrium. They are said to be at the same temperature.  The Concept of Temperature and Heat Definition of temperature There is a scalar quantity called temperature that assumes the same value for bodies in thermal equilibrium.  The Zeroth Law of Thermodynamics Let’s now introduce a third body C ZEROTH LAW OF THERMODYNAMICS If bodies A and B are separately in thermal equilibrium with a third body C, then bodies A and B will be in thermal equilibrium if they are placed in thermal contact.  The Zeroth Law of Thermodynamics What is the third body C? The third body is a thermometer. A thermometer is an instrument that measures the A and C are in B and C are in temperature of a body thermal equilibrium thermal equilibrium in a quantitative way. Temperature is the only factor that determines whether two objects in A and B are in thermal contact are in thermal equilibrium thermal equilibrium or not.  Thermal Energy (Heat) versus Temperature Thermal Energy (Heat) Temperature is kinetic energy in transit from one is the average kinetic energy of object to another due to temperature particles in an object – not the total difference. amount of kinetic energy particles. Measured in calories or Joules Measured in C, F, K Any object with temperature above Absolute zero is zero Kelvin zero Kelvin has thermal energy 𝑇1 𝑇2 Heat  Thermodynamics The study of the effects of work, heat, and energy on a system A thermodynamic system is a definite quantity of matter enclosed by boundaries, either real or imaginary. Everything outside the boundary is called the surroundings. The system may interact with the surroundings by interchanging energy through a transfer of heat and/or the performance of mechanical work.  Thermal energy flows from hot to cold Identify the system and the surroundings Indicate the direction of energy flow  Thermodynamic Systems Open: Matter CAN cross the boundary Open matter system heat Closed: Matter CANNOT cross the boundary Closed system heat Isolated: Boundary seals matter and heat from being Isolated exchanged system  Work – Mechanical Energy Transfer Work done by expanding gas (small change in volume) 𝑊 = −𝐹Δ𝑠 = 𝑃𝐴Δ𝑠 = 𝑃Δ𝑉  Work – Mechanical Energy Transfer Work done depends on the thermodynamic process. 𝑊𝐴𝐵𝐶𝐷𝐴 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 𝐴𝐵𝐶𝐷𝐴 𝑊𝐴𝐶𝐷𝐴 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝐴𝐶𝐷𝐴 Therefore, 𝑊𝐴𝐵𝐶𝐷𝐴 ≠ 𝑊𝐴𝐶𝐷𝐴 Similarly, heat absorbed or rejected by a system depends on the thermodynamic process. Internal Energy of a Thermodynamic System The internal energy of a system is the sum of the kinetic and potential energies of the atoms and molecules making up the system.  The First Law of Thermodynamics The first law of thermodynamics is a statement of conservation of energy. If a system’s volume is constant, and heat is added, its internal energy increases.  The First Law of Thermodynamics If a system does work on the external world, and no heat is added, its internal energy decreases.  The First Law of Thermodynamics Combining these gives the first law of thermodynamics. It is vital to keep track of the signs of Q and W.  The First Law of Thermodynamics The internal energy of the system depends only on the state of the system (P, V, T), not on the process. The work done and the heat added, however, depend on the details of the process involved.  Example In the figure, the gas absorbs 400J of heat and at the same time does 120 J of work on the piston. What is the change in the internal energy of the system? SOLUTION By the first law of thermodynamics Δ𝑈 = 𝑄 − 𝑊 = +400 𝐽 − +120 𝐽 = +280 𝐽  Processes in Thermodynamics Isothermal process → takes place at constant temperature (e.g. freezing of water to ice at 0°C) Any heat flow into or out of the system must be slow enough to maintain thermal equilibrium For ideal gases, 𝑈 = 32 𝑁𝑘𝑇 = 32 𝑛𝑅𝑇 𝑤ℎ𝑒𝑟𝑒, 𝑁 ≡ Number of particles, 𝑘 ≡ Boltzmann constant = 1.3806 × 10−23 JK −1 and 𝑅 ≡ gas constant = 8.314 J. mol−1 K −1 For isothermal processes Δ𝑇 is zero, and, hence, Δ𝑈 = 0 Therefore, Q = W by the First Law Any energy entering the system (Q) must leave as work (W) For an ideal gas undergoing an isothermal process, pressure varies inversely with volume. 𝑛𝑅𝑇 𝑉𝑓 𝑃= Q = W = 𝑛𝑅𝑇 ln 𝑉 𝑉𝑖 𝑻 = 𝒄𝒐𝒏𝒔𝒕. Q=W>0 Processes in Thermodynamics Isobaric process → takes place at constant pressure (e.g. heating of water in open air under atmospheric pressure) ΔU, W, and Q are generally non- zero, but calculating the work done by an ideal gas is straightforward 𝑊 = 𝑃Δ𝑉  Processes in Thermodynamics Isochoric process → takes place at constant volume (e.g. heating of gas in a sealed metal container) When the volume of a system doesn’t change, it will do no work on its surroundings. 𝑊 = 0 Δ𝑈 = 𝑄  Processes in Thermodynamics Adiabatic process (No heat transfer) System is thermally insulated or undergoes a fast process Therefore 𝑄 = 0 and Δ𝑈 = – 𝑊 When a system expands adiabatically, 𝑊 is positive (the system does work) so Δ𝑈 is negative. When a system compresses adiabatically, 𝑊 is negative (work is done on the system) so Δ𝑈 is positive. Processes in Thermodynamics The adiabatic P-V curve is similar to the isothermal one, but is steeper. The P-V curve for an adiabat is given by  𝐹𝑜𝑟 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑐ℎ𝑎𝑛𝑔𝑒 𝑊𝑎𝑑𝑖𝑎𝑏𝑎𝑡 < 𝑊𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚 For an ideal gas: 𝑊𝑎𝑑𝑖𝑎𝑏𝑡 = Δ𝑈 = 𝐶𝑉 𝑇2 − 𝑇1 Or 𝑊 𝑃2 𝑉2 − 𝑃1 𝑉1 𝑊𝑎𝑑𝑖𝑎𝑏𝑎𝑡 = 𝛾−1 In summary:  Other Processes in Thermodynamics Reversible process → system is close to equilibrium at all times Very slow processes are considered equilibrium processes Cyclic process → the final and initial state are the same. For the ideal gas, Δ𝑈 = 0 and 𝑄 = 𝑊 Internal Energy in closed cycles In closed cycles State A: PA, VA, TA, UA ΔU=0 Internal energy does not depend on the process. It is a state function Work and Heat depend on the process. They are not State B: PB, VB, TB, UB state functions  Specific Heats for an Ideal Gas Specific heats for ideal gases must be quoted either at constant pressure or at constant volume. For a constant-volume process, Where 𝐶𝑉 is the molar heat capacity at constant volume. Molar heat capacity C is defined as the heat per unit mole per Celsius degree: 𝑄 𝐶= 𝑛Δ𝑇 Specific Heats for an Ideal Gas Example How much heat is required At constant pressure, to raise the temperature of 2 moles of O2 from 0oC to 100oC? 𝐶𝑃 = 29.5 𝐽Τ𝑚𝑜𝑙. 𝐾 𝐶𝑉 = 21.1 𝐽Τ𝑚𝑜𝑙. 𝐾 SOLUTION 𝑄𝑃 = 𝑛𝐶𝑃 Δ𝑇 𝑄𝑃 = 2 × 29.5 × 100 𝑄𝑃 = +5900 𝐽 𝑄𝑉 = 𝑛𝐶𝑉 Δ𝑇 𝑄𝑉 = 2 × 21.1 × 100 𝑄𝑉 = +4220 𝐽  Specific Heats for an Ideal Gas (PV = nRT) Both CV and CP can be calculated for a monatomic ideal gas using the first law of thermodynamics. 3 3 For an Idea gas: 𝑈 = 𝑛𝑅𝑇, 𝑆𝑜 Δ𝑈 = 𝑛𝑅Δ𝑇 2 2 From the first law: Δ𝑈 = 𝑄 − 𝑊 3 For small Δ𝑉, the first law becomes: 𝑛𝑅Δ𝑇 = 𝑄 − 𝑃Δ𝑉 2 𝑄 3 𝑃Δ𝑉 Dividing by nΔ𝑇 and rearranging: 𝐶 = = 𝑅+ 𝑛Δ𝑇 2 𝑛Δ𝑇 3 𝑛𝑅Δ𝑇 5 At constant pressure, 𝐶𝑃 = 𝑅 + = 𝑅 (𝑢𝑠𝑖𝑛𝑔 𝑃𝑉 = 𝑛𝑅𝑇) 2 𝑛Δ𝑇 2 3 At constant volume: 𝐶𝑉 = 𝑅 2  Specific Heats for an Ideal Gas Although the calculation in the previous slide was done for an ideal, monatomic gas, it works well for real gases.  Calculating Energy Transfer Calculate the energy transferred when a block of aluminum at 80.0°C is placed in 1.00 liter (1kg) of water at 25.0°C if the final temperature Q becomes 30.0°C. Step 1. List all known values Mass of water = 1 kg Cp of water = 4184 JΤkg°C Difference in temperature (water) ΔT = 30.0°C – 25.0°C = 5.0°C Cp of Al = 900 JΤkg°C Difference in temperature (Al) ΔT = 80.0°C – 30.0°C = 50.0°C Calculating Energy Transfer Step 2. List all unknown values 𝑄 = 𝑒𝑛𝑒𝑟𝑔𝑦 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑟𝑒𝑑 𝑚𝐴𝑙 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝐴𝑙 𝑏𝑙𝑜𝑐𝑘 Step 3. Select equations to solve unknown values 𝑄 = 𝑚𝐶𝑝 𝛥𝑇 𝑄𝐴𝑙 = 𝑄𝑤𝑎𝑡𝑒𝑟 Step 4. Solve for Qwater Qwater = 1.00kg × 4184 𝐽Τ𝑘𝑔 ℃ × 5.0℃ = 20,920 J gained Calculating Energy Transfer Step 5. Solve for 𝑚𝐴𝑙 QAl (lost) = Qwater (gained) = 20,920 J 𝑄𝐴𝑙 = 𝑚𝐴𝑙 𝐶𝑝 𝛥𝑇 QAl 20920J mAl = = Cp ΔT 900 𝐽Τ𝑘𝑔 ℃ × 50.0℃ 𝑚𝐴𝑙 = 0.465 kg = 465g  Example C A 2-L sample of Oxygen gas has an initial temperature of 200 K and pressure of 1 atm. The gas undergoes D four processes: AB: Heated at constant V to 400 K. BC: Heated at constant P to 800 K. SOLUTION CD: Cooled at constant V back to 1 atm. 𝑃𝑖 𝑉𝑖 1.01 × 105 𝑃𝑎 × 0.002𝑚3 DA: Cooled at constant P back to 200 K. 𝑛= = 𝑅𝑇𝑖 8.314 𝐽Τ𝑚𝑜𝑙. 𝐾 × 200𝐾 1. How many moles of 𝑂2 are present? 𝑛 = 0.121 𝑚𝑜𝑙  Example C A 2-L sample of Oxygen gas has an initial temperature of 200 K and pressure of 1 atm. The gas undergoes D four processes: AB: Heated at constant V to 400 K. BC: Heated at constant P to 800 K. SOLUTION CD: Cooled at constant V back to 1 atm. 𝑛𝑅𝑇𝐵 𝑃𝐵 = DA: Cooled at constant P back to 200 K. 𝑉𝐵 0.121 𝑚𝑜𝑙×8.314𝐽Τ𝑚𝑜𝑙.𝐾×400𝐾 𝑃𝐵 = 0.002𝑚3 2. What is the pressure at point B? 𝑃𝐵 = 2.02 × 105 𝑃𝑎 = 2 𝑎𝑡𝑚  Example C A 2-L sample of Oxygen gas has an initial temperature of 200 K and pressure of 1 atm. The gas undergoes D four processes: AB: Heated at constant V to 400 K. BC: Heated at constant P to 800 K. SOLUTION CD: Cooled at constant V back to 1 atm. 𝑄𝐴𝐵 = 𝑛𝐶𝑉 Δ𝑇𝐴𝐵 DA: Cooled at constant P back to 200 K. 𝑄 = 0.121 𝑚𝑜𝑙 × 21.1 𝐽Τ𝑚𝑜𝑙. 𝐾 × 200 𝐾 𝑄𝐴𝐵 = 511 𝐽 Δ𝑈𝐴𝐵 = 𝑄𝐴𝐵 = 511 𝐽 3. Calculate Q, W and ΔU for the process AB. 𝑊 = 0 (Isochoric Process)  Example C A 2-L sample of Oxygen gas has an initial temperature of 200 K and pressure of 1 atm. The gas undergoes D four processes: AB: Heated at constant V to 400 K. BC: Heated at constant P to 800 K. SOLUTION CD: Cooled at constant V back to 1 atm. 𝑃𝐶 = 𝑃𝐵 = 203 𝑘𝑃𝑎 DA: Cooled at constant P back to 200 K. 𝑛𝑅𝑇𝐶 𝑉𝐶 = 𝑃𝐶 0.121 𝑚𝑜𝑙×8.314 𝐽Τ𝑚𝑜𝑙.𝐾×800 𝐾 𝑉𝐶 = 4. What is the volume at point C (and D)? 202000 𝑃𝑎 𝑉𝐶 = 0.004 𝑚3 = 4 𝐿  Example C A 2-L sample of Oxygen gas has an initial temperature of 200 K and pressure of 1 atm. The gas undergoes D four processes: AB: Heated at constant V to 400 K. BC: Heated at constant P to 800 K. SOLUTION CD: Cooled at constant V back to 1 atm. DA: Cooled at constant P back to 200 K. Δ𝑈𝐵𝐶 = 𝑛𝐶𝑉 Δ𝑇𝐵𝐶 Δ𝑈𝐵𝐶 = 0.121 𝑚𝑜𝑙 × 21.1 𝐽Τ𝑚𝑜𝑙. 𝐾 × 400𝐾 5. What is the change in U along BC? Δ𝑈𝐵𝐶 = 1021 𝐽  Example C A 2-L sample of Oxygen gas has an initial temperature of 200 K and pressure of 1 atm. The gas undergoes D four processes: AB: Heated at constant V to 400 K. BC: Heated at constant P to 800 K. SOLUTION CD: Cooled at constant V back to 1 atm. 𝑊𝐵𝐶 = 𝑃𝐵 Δ𝑉𝐵𝐶 DA: Cooled at constant P back to 200 K. 𝑊𝐵𝐶 = 2.02 × 105 𝑃𝑎 × 4 𝐿 − 2 𝐿 = 404 𝐽 Q BC = Δ𝑈𝐵𝐶 + 𝑊𝐵𝐶 6. Find W and Q in the process BC. 𝑄𝐵𝐶 = 1023𝐽 + 404 𝐽 = 1427 𝐽  Example C A 2-L sample of Oxygen gas has an initial temperature of 200 K and pressure of 1 atm. The gas undergoes D four processes: AB: Heated at constant V to 400 K. BC: Heated at constant P to 800 K. SOLUTION CD: Cooled at constant V back to 1 atm. 𝑇𝐷 𝑃𝐷 𝑃𝐷 DA: Cooled at constant P back to 200 K. = ⇒ 𝑇𝐷 = 𝑇𝐶 𝑇𝐶 𝑃𝐶 𝑃𝐶 1 𝑎𝑡𝑚 𝑇𝐷 = × 800 𝐾 = 400 𝐾 7. What is temperature in state D? 2 𝑎𝑡𝑚  Example C A 2-L sample of Oxygen gas has an initial temperature of 200 K and pressure of 1 atm. The gas undergoes D four processes: AB: Heated at constant V to 400 K. BC: Heated at constant P to 800 K. SOLUTION CD: Cooled at constant V back to 1 atm. 𝑄𝐶𝐷 = 𝑛𝐶𝑉 Δ𝑇 DA: Cooled at constant P back to 200 K. 𝑄𝐶𝐷 = 0.121 𝑚𝑜𝑙 × 21.1 𝐽Τ𝑚𝑜𝑙. 𝐾 × 400 − 800 𝐾 𝑄𝐶𝐷 = −1021 𝐽 8. Find Q and ΔU in the process CD? Δ𝑈𝐶𝐷 = 𝑄𝐶𝐷 = −1021 𝐽 W=0  Example C A 2-L sample of Oxygen gas has an initial temperature of 200 K and pressure of 1 atm. The gas undergoes D four processes: AB: Heated at constant V to 400 K. BC: Heated at constant P to 800 K. SOLUTION 𝑊𝐷𝐴 = 𝑃𝐴 ΔVDA CD: Cooled at constant V back to 1 atm. 𝑊𝐷𝐴 = 1.01 × 105 𝑃𝑎 × 0.002 − 0.004 𝑚3 DA: Cooled at constant P back to 200 K. 𝑊𝐴𝐵 = −202 𝐽 Δ𝑈𝐷𝐴 = 𝑛𝐶𝑉 Δ𝑇𝐷𝐴 Δ𝑈𝐷𝐴 = 0.121 𝑚𝑜𝑙 × 21.1 𝐽Τ𝑚𝑜𝑙. 𝐾 × −200 𝐾 Δ𝑈𝐷𝐴 = −511 𝐽 9. Find Q, W and ΔU in the process DA? 𝑄𝐶𝐷 = Δ𝑈𝐷𝐴 + 𝑊𝐷𝐴 = −511 − 202 𝐽 𝑄𝐷𝐴 = −713 𝐽  Example A 2-L sample of Oxygen gas has an initial temperature C of 200 K and pressure of 1 atm. The gas undergoes four processes: D AB: Heated at constant V to 400 K. BC: Heated at constant P to 800 K. SOLUTION CD: Cooled at constant V back to 1 atm. 𝑊𝑐𝑦𝑐𝑙𝑒 = 𝑊𝐴𝐵 + 𝑊𝐵𝐶 + 𝑊𝐶𝐷 + 𝑊𝐷𝐴 DA: Cooled at constant P back to 200 K. 𝑊𝑐𝑦𝑐𝑙𝑒 = 0 + 404 𝐽 + 0 − 202 𝐽 = 202 𝐽 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑙𝑜𝑜𝑝 𝐴𝐵𝐶𝐷𝐴 = (𝑃𝐵 − 𝑃𝐴 )(𝑉𝐶 − 𝑉𝐵 ) 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑙𝑜𝑜𝑝 𝐴𝐵𝐶𝐷𝐴 = 1 𝑎𝑡𝑚 × 2 𝐿 = 202 𝐽 10. Check that ΣΔU = 0 for the cycle Δ𝑈𝑐𝑦𝑐𝑙𝑒 = Δ𝑈𝐴𝐵 + Δ𝑈𝐵𝐶 + Δ𝑈𝐶𝐷 + Δ𝑈𝐷𝐴 and total W = area of loop ABCDA. Δ𝑈𝑐𝑦𝑐𝑙𝑒 = 511 𝐽 + 1021 𝐽 − 1021 𝐽 − 511 𝐽 = 0 More Examples and Quizzes Water has a specific heat of 1 cal/gmK and iron has a specific heat of 0.107 cal/gmK. If we add the same amount of heat to equal masses of iron and water, which will have the larger change in temperature? A.The iron. B. They will have equal changes since the same amount of heat is added to each. C. The Water. D.None of the above Compare the amount of heat energy required to raise the temperature of 1 kg of water and 1 kg of iron 20C. 𝑄 = 𝑚𝐶Δ𝑇 For Water 𝑄 = (1000𝑔)(1𝑐𝑎𝑙/𝑔°𝐶)(20°𝐶) = 20,000𝑐𝑎𝑙 For Iron 𝑄 = (1000𝑔)(0.107𝑐𝑎𝑙/𝑔°𝐶)(20°𝐶) = 2140𝑐𝑎𝑙 A cylinder of radius 5 cm is kept at constant pressure with a piston of mass 75 kg. a) What is the pressure inside the cylinder? 1.950x105 Pa b) If the gas expands such that the cylinder rises 12.0 cm, what work was done by the gas? 183.8 J c) What amount of the work went into changing the gravitational PE of the piston? 88.3 J d) Where did the rest of the work go? Compressing the outside air Identify the processes P Isobaric P = constant V P Isochoric V = constant V W = 0 P Isothermal V T = constant P U = 0 (ideal gas) V Adiabatic Q = 0 a) What amount of work is performed by the gas in the cycle IAFI? W=304 J b) How much heat was absorbed by the gas in the cycle IAFI? Q = 304 J c) What amount of work is performed by the gas in the cycle IBFI? W = -304 J Consider a monotonic ideal gas. a) What work was done by the gas from A to B? 20,000 J b) What heat was added to P (kPa) the gas between A and B? 75 A 20,000 J c) What work was done by the gas from B to C? 50 -10,000 J d) What heat was added to B the gas beween B and C? 25 -25,000 J C e) What work was done by V (m3) the gas from C to A? 0 f) What heat was added to 0.2 0.4 0.6 the gas from C to A? 15,000 J Take solutions from last problem and find: a) Net work done by gas in the cycle b) Amount of heat added to gas WAB + WBC + WCA = 10,000 J QAB + QBC + QCA = 10,000 J Consider an ideal gas undergoing the path through the PV diagram. In going from A to B to C, (a) the work done BY the gas is _______ 0. (b) the change of the internal energy of the gas is _______ 0. (c) the amount of heat added to the gas is _______ 0. a) > b) < P C c) = B A V In going from A to B to C to D to A, (a) the work done BY the gas is _______ 0. (b) the change of the internal energy of the gas is _______ 0. (c) the heat added to the gas is _______ 0. a) > P D C b) < c) = B A V  Modes of Heat Transfer Heat transfer is exchange of thermal energy due to a difference in temperature. Modes Conduction Heat transfer through a solid or stationary medium due to molecular collisions. Convection Heat transfer due to the bulk movement of fluid. Radiation Heat transfer via electromagnetic waves, requiring no medium.  Modes of Heat Transfer The first law of thermodynamics does not provide the equations for heat transfer 𝑄= Δ𝑈 ด + 𝑊 ณ 𝑐𝑎𝑛 𝑏𝑒 𝑑𝑒𝑡𝑒𝑟𝑛𝑖𝑛𝑒𝑑 𝑐𝑎𝑛 𝑏𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑖𝑓 𝑡ℎ𝑒 𝑠𝑡𝑎𝑡𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑐𝑒 𝑑𝑜𝑖𝑛𝑔 𝑠𝑦𝑠𝑡𝑒𝑚 𝑖𝑠 𝑘𝑛𝑜𝑤𝑛: 𝑤𝑜𝑟𝑘 𝑖𝑠 𝑘𝑛𝑜𝑤𝑛: (𝑃,𝑉,𝑇,𝑛) 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑠𝑝𝑟𝑖𝑛𝑔 𝑤𝑜𝑟𝑘, 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑤𝑜𝑟𝑘, 𝑒𝑡𝑐. We need equations for heat transfer!  Conduction Conduction is a mechanism of heat transfer through a solid or between solids in direct contact. Thermal energy is transferred from more energetic to less energetic particles due to their interactions Requires a temperature gradient. no movement of the material itself. Example: Heat conduction through a metal rod  Conduction Key Formula for conduction Fourier’s Law of Heat Conduction Δ𝑇 𝑞 = −𝑘𝐴 Δ𝑥 Where: 𝑞: Heat transfer rate (W) 𝑘: Thermal conductivity (W/m.K) 𝐴: Cross-sectional area (m²) Δ𝑇 : Temperature gradient (K/m) Δx  Conduction Example: Heat Transfer through a Wall A wall 0.2 m thick has a thermal conductivity of 0.8 W/m.K. The temperatures on the two sides are 20°C and 5°C. The wall area is 10 m². Solution: 𝑘𝐴(𝑇1 − 𝑇2 ) 𝑞= 𝐿 Substituting values: 0.8 × 10 × (20 − 5) 𝑞= = 600W 0.2  Conduction Example: Heat Transfer through a Wall A cubical freezer compartment is 2 m on a side, with the bottom is perfectly insulated. The thermal conductivity of the styrofoam insulation is 0.030 W/ m. K. The temperatures inside and outside the cube are, respectively, − 10°C and 35°C. What should the thickness of the styrofoam insulation be if the heat load is 500W? Solution: 𝑘𝐴(𝑇1 − 𝑇2 ) 2m 𝐿= 𝑞 2m Substituting values: 2m 0.030 × (5 × 22 ) × (35 − (−10)) 𝐿= = 0.054𝑚 500  Convection Convection is Heat transfer through the movement of fluid (liquid or gas). Types: Natural Convection: Driven by buoyancy due to temperature differences. Example: Cooling of a hot cup of coffee in air. Forced Convection: Driven by external forces like fans or pumps. Example: Cooling of PC by fan.  Convection Key Formula for Convection ℎ depends on Newton’s Law of Cooling fluid property (T,P) 𝑞 = ℎ𝐴(𝑇𝑠 − 𝑇∞ ) Geometry (flat, circular, sphere) flow regime (laminar, turbulent) etc. Where: ℎ = 25 t0 250 W/m2K for gases 𝑞: Heat transfer rate (W) ℎ = 100 t0 20000 W/m2K liquids ℎ: Convective heat transfer coefficient (W/m²·K) 𝐴: Surface area (m²) 𝑇𝑠 : Surface temperature (K or °C) 𝑇∞ : Surrounding fluid temperature (K or °C)  Convection Example: Cooling of a Heated Plate A plate at 100°C is exposed to air at 25°C. The surface area is 2 m², and the convective heat transfer coefficient is 10 W/m²·K. Solution: 𝑞 = ℎ𝐴(𝑇𝑠 − 𝑇∞ ) = 10 × 2 × (100 − 25) = 1500W  Convection Example: Cooling of Electric Resistance An electric resistance heater is embedded in a large cylinder of diameter 30mm and surface temperature 90°C. Air at 23°C flows over the cylinder. The heater puts out 400 W/m of thermal energy. What is the value of the convective heat transfer coefficient, ℎ? Solution: 𝑇𝑆 , ℎ 𝑞 𝑞 𝑞/𝐿 Air at 𝑇∞ ℎ= = = 𝐴 𝑇𝑠 − 𝑇∞ 𝜋𝐷𝐿 𝑇𝑠 − 𝑇∞ 𝜋𝐷 𝑇𝑠 − 𝑇∞ 400 ℎ= = 65W/m2. K 𝜋 × 0.030 × (90 − 23)  Radiation Radiation is Heat transfer through electromagnetic waves does not require a medium. depends on temperature and surface properties. Example Heat from the Sun reaching Earth.  Radiation Key Formula for Radiation Stefan-Boltzmann Law 𝑞 = 𝜎𝐴𝑇 4 𝑏𝑙𝑎𝑐𝑘𝑏𝑜𝑑𝑦 𝑟𝑎𝑑𝑖𝑎𝑖𝑜𝑛 (𝑖𝑑𝑒𝑎𝑙 𝑒𝑚𝑖𝑡𝑡𝑒𝑟) 𝑞 = 𝜎𝜖𝐴𝑇 4 for a non−blackbody Where: 𝑞: Radiative heat transfer rate (W) 𝜎: Stefan-Boltzmann constant (5.67 × 10−8 W/m2. K 4 ) 𝜖: Emissivity of the surface (dimensionless, 0 ≤ 𝜖 ≤ 1) 𝜖 = 1 for a blackbody 𝐴: Surface area (m²) 𝑇: Surface temperature (K)  Radiation Key Formula for Radiation Stefan-Boltzmann Law For a small body in an enclosure (surrounding) 𝑞 = 𝜎𝜖𝐴 𝑇 4 − 𝑇∞4 Emissive power: 𝑞 𝐸= W/m2 Where 𝑇∞ ​: Surrounding temperature (K) 𝐴 Absorbed incident radiation: 𝐺𝑎𝑏𝑠 = 𝛼𝐺 𝑇 𝑇∞ Where: 𝐺 ≡ 𝑖𝑟𝑟𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛 𝑊/𝑚2 𝛼 ≡ 𝑎𝑎𝑏𝑠𝑜𝑟𝑏𝑡𝑖𝑣𝑖𝑡𝑦 0≤𝛼≤1  Radiation Example: Heat Loss from a Radiating Surface A surface at 500 K radiates heat to surroundings at 300 K. The surface area is 1 m², and the emissivity is 0.9. Solution: 𝑞 = 𝜎𝜖𝐴(𝑇 4 − 𝑇∞4 ) Substituting values: 𝑞 = 5.67 × 10−8 × 0.9 × 1 × 5004 − 3004 = 2776W  Radiation Example: Heat Loss from a Radiating Surface A spherical instrumentation package has a diameter of 100mm, with 𝜖 = 0.023 in a large space simulation chamber whose walls are at 77 K the outside temperature of the package is 40°C. How much power is being dissipated by the package?. Solution: 𝑞 = 𝜎𝜖𝐴(𝑇 4 − 𝑇∞4 ) Substituting values: 𝑞 = 5.67 × 10−8 × 0.023 × 𝜋 0.1 2 × 3134 − 774 = 4.30W  End of Chapter 9