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Thermodynamics Ch 16 – Temperature and Heat 16.1 Temperature and the Zeroth Law of Thermodynamics The Zeroth Law of Thermodynamics If two objects are separately in thermal equilibrium with a 3rd object, then the first two objects are in thermal equilibrium with each other. Thermal equilibrium: If two objects are in thermal equilibrium then neither object will exchange energy by heat or electromagnetic radiation if they are placed in thermal contact. Temperature is a property that determines whether an object is in thermal equilibrium. Thermal contact: If in thermal contact two objects will exchange energy between them if a temperature difference. 16.2 Temperature Scales You shouldn’t ever have to memorize formulas in Physics. Take for example the formula from or to Celsius to/from Fahrenheit. WATER Freezing Boiling Difference Fahrenheit 32° 212° 180° Celsius 0° 100° 100° Let’s try using this logic… Take the ratio of 180/100 = 9/5 & 100 / 180 = 5/9 98.6 °F Where is the “zero” (with respect to water) for the Let’s bring the Fahrenheit scale down to a Celsius scale? more logical “zero” point. Ans: 0 °C 98.6 °F - 32 °F = 66.6 F° (take note of the units, 66.6 F°) For the Fahrenheit scale? Ans:32 °F We now need to bring this “BIG” scale down Now, which scale has the most number between to the “small” scale by multiplying by 5/9 freezing and boiling? 66.6 F° * 5/9 = 37 °C Fahrenheit (we’ll call this the “BIG” scale; conversely Celsius is the “small” scale) Let’s go in reverse… So we say the “zero” point for the Fahrenheit scale is 32 °F… 37 °C uses the “small” scale I think most people will agree this is a weird let’s go to the “BIG” scale “zero” point 37 °C * 9/5 = 66.6 F° How can we bring this “zero” point to zero? o Ans: Subtract 32 °F But the “zero” point of the Fahrenheit scale We also recognize that the Fahrenheit scale is the isn’t “zero”…it’s 32 °F “BIG” scale…so how can we bring it to a small 66.6 F° + 32 F° = 98.6 °F scale? o Ans: Multiply by a small ratio, 5/9 The Kelvin scale is the proper scale in most situations (PV = nRT, etc). When in doubt…just use the Kelvin scale. Why? 1 Kelvin = 1 °C and T = TC + 273.15 So if the temps are given in °C then ΔT will be the same whether in the Kelvin or Celsius scales. If you notice T temperature in the Kelvin scale TC temperature in Celsius TF temperature in Fahrenheit The Kelvin scale is an absolute scale which means you can not go below zero. At 0 K (do not use the degree sign) all movement (i.e. in an atom) stops. As you can see, we can not achieve the Absolute Zero point, but we can achieve (optional) Thermometers make use of thermometric properties. Types of thermometers volume expansion (Hg column) constant volume gas (Δ gas pressure) thermocouples (Voltage differences of two different metals) electrical resistance (Platinum wire) temps very close. thermographic (emitted radiation) Galileo’s Thermometer Demo: TH-E-GT 16.3 Thermal Expansion (of Solids and Liquids) Let’s pretend we have a 10 cm steel ruler an a 2 meter stick (also steel) Now let’s pretend we have two 10 cm steel rulers, A and B. If we heat both of the steel sticks…which will expand more a greater amount? 10 cm steel ruler, A, undergoes ΔTC = 10 C° 10 cm steel ruler, B, undergoes ΔTC = 100 C° Which will expand more? Most of you will say the long one will expand more and you’ll be RIGHT! Some may say they will expand the same and you’ll be CORRECT also (almost)! Explanation: Since they are both steel, they will expand by the same proportion; but not the same amount… (We know the longer one will expand by the greater total length since 5% of 2 meters is much greater than 5% of 10 centimeters) B, will expand a factor of 10 greater A Most of you already knew these results. These results also lead to one conclusion: ΔL = α L ΔT Which states that the change of length of an object is dependant on Type of material Initial Length of material Change of temperature Application of Thermal Expansion/Contraction You are attempting to What happens to the coupling? form the best junction it expands, including the hole between two pipes. in the middle What happens to the pipes? Question: What would happen if the coupling and the pipes were made of different materials and the entire new system became very hot or very cold? Use a coupling between pipes heat the coupling (torch) cool the ends of the pipes (ice bath or oil bath cooled with liquid nitrogen) they contract, become smaller Ans: Different materials have different coefficients of linear expansion, so if the coupling is a Slide the pipes into the different material, the coupling would coupling…let achieve room temp. either crack (contract too much) or leak You now have a coupling as strong (expand too much) as the pipe itself. This is the same principle as the bimetallic strip discussed in your book. Volume Thermal Expansion One of the notable exceptions to First NOTE: β equals 3 * α Coef of Volume Expansion = 3*(Coef of Linear Exp) materials contracting as they get colder. WATER!!! Water’s highest density is at 4 °C This is very important for life, if you don’t see why…ask in CLASS!!! Derivation below And the formula follows similarly to linear expansion ΔL = α L ΔT ΔV = β V ΔT Vo + V Vo + V Vo + V = (l + l) (h + h) (w + w) = (l + lT) (h + h T) (w + w T) = l*h*w (1 + T)3 Derivation: Remember Young’s Modulus…we now have another application Young’s Modulus = Stress / Strain; where stress F/A and strain is ΔL/L (Y) (Y) ΔL/L Stress Stress Stress = F/A / ΔL/L = F/A (= stress) = (Y) ΔL /L ; where ΔL = α L ΔT = (Y) α L ΔT / L = Y α ΔT So if you know Young’s Modulus and the coef of linear expansion, you can calculate Stress. Special Properties of water Vo + V V/Vo = Vo (1 + 3T + 3(T)2 + (T)3) = 3T + 3(T)2 + (T)3 We also know T <<1 for most materials…so V/Vo = 3T V = Vo3T Thermostat Demo: TH-A-DT 16.4 Heat and Mechanical Work Units of heat calorie Heat is defined as the transfer of energy across the boundary of a system due to a temperature difference between the system and its surroundings. the amount of energy transfer required to raise the temp of 1 gram of water from 14.5 °C to 15.5 °C joule The SI unit: 1 cal = 4.186 Joules 1 kcal = 1 Calorie = 4186 Joules BTU English unit: (Q required to raise 1 lb, 1 °F) 1 Btu = 1054 Joules If work is done on a system, then the system will transfer energy across its boundary. Work on system The dentists drill causes your teeth to get hot…so cold water is sprayed on your teeth to cool down the teeth. The cold water is external to the system and its temperature raises when in contact with the hot tooth. 16.5 Specific Heats Work by system If energy is added to a system, work may be done by the system. Burn coal in a steam engine, and the steam engine is able to move the entire train. Heat Capacity, C : Q = C ΔT C, the heat capacity is defined as the energy needed to raise the temperature of a sample by 1°C Specific Heat: Q = m c ΔT c, the heat capacity of a sample per unit mass Question: Which would require more energy to bring to boil if both are initially at room temperature? A) A cup of water B) A swimming pool full of water Question: Which would require more energy to rise from 20 °C to 50 °C? A) A cup of water B) A cup of alcohol Ans: I wouldn’t want to have to pay the energy bill required to raise a swimming pool full of water to its boiling point. Ans: This one is harder…but I believe many know raising the temperature of alcohol requires less heat than water. Conclusion: The greater the mass…the more energy required to bring to boiling point. Question: Which would require more energy to bring to boil? A) A cup of water at 20 °C B) A cup of water at 90 °C Conclusion: The material being heater makes a bit difference, so materials like water require much more energy to change its temperature compared to materials like alcohol and sand. Question: Which would require more energy to bring to 20 °C? C) A cup of ice at 0 °C D) A cup of water at 0 °C Ans: This one is easy…the one at 90 °C is almost at boil already, thus the energy input required to bring it to boil will way less than the cup at 20°C Ans: This one is easy…imagine adding 0°C water to your Coke or Pepsi to cool it down. 0°C ice requires much energy to bring it to 0°C water. Conclusion: The greater the ΔT…the greater amount of energy was required. Conclusion: The change of state requires a large change of energy. SI units are generally preferable, but I’m presenting this section in cgs units Question: A large block of ice, 1000 g, is at -20 °C. How much energy is required to bring it to 100 °C steam? Ans: Q Q Q Q Q Q v vaporization f fusion units of c calories / (gram C°) = cice m ΔT + m Lf + cwaterm ΔT + m Lv = ½ (1000)(20) + 1000(80) + 1 (1000)(100) + 1000 (539) = 10,000 cal + 80,000 cal + 100,000 cal + 539,000 cal = 729,000 calories (1 Calorie = 1 kilocalorie) = 729 Calories (4.186 J / Cal) = 3052 Joules (a good exercise for you is to work the problem out using 16.6 Conduction, Convection, Radiation Conduction P=Q/t P = kA ΔT/Δx P = kA ΔT/Δx P = A ΔT/(Δx/k) P = A ΔT/(ΣRi) Δx: The thicker the wall…more isolated A: More area… the more energy flows through the wall ΔT: The greater the temperature difference… the more energy flow through the wall k: And different materials will have different coefficients for heat flow R ft2 °F hr/Btu 1 ft2 °F hr/Btu (0.305m/ft)2 (5°C/9°F)(3600s/hr)(1Btu/1054J) 1 ft2 °F hr/Btu = 0.1765 m2 °F sec / J Standard 2x4 insulation is R-11 English units In metric this would be R-2 Radiation Convection Stefan’s Law: P = σAeT4 SI units) Energy transferred by the movement of a warm medium Convection Currents in Air: TH-B-CC σ = 5.669 x 10-8 W/m2K4, e is emissivity (0 to 1) Ideal absorber, e = 1 (Black Body) Ideal reflector, e = 0 Demo: Light the Match: TH-B-LM Crooke’s Radiometer Demo: TH-D-CR