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Physics 1710 —Warm-up Quiz

Physics 1710 —Warm-up Quiz. Answer Now !. 0. 35% 49 of 140. 0. In Antarctica the thermometer reads –20 C; what is this temperature on a Fahrenheit thermometer?. -36 o F -11 o F -4 o F -43 o F -20 o F. 0. Physics 1710 C hapter 20 Heat & 1 st Law of Thermo. Solution:

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Physics 1710 —Warm-up Quiz

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  1. Physics 1710—Warm-up Quiz Answer Now ! 0 35% 49 of 140 0 In Antarctica the thermometer reads –20 C; what is this temperature on a Fahrenheit thermometer? • -36 oF • -11 oF • -4 oF • -43 oF • -20 oF

  2. 0 Physics 1710 Chapter 20 Heat & 1st Law of Thermo Solution: oF = (180 oF/100 oC) oC +32oF =(1.8 oF/oC)(-20 oC) +32oF = (-36 +32) oF = -4oF

  3. No Talking! Think! Confer! 0 What will happen to a heated ring? The hole will(1) get larger; (2) get smaller; (3) stay the same. Physics 1710 Chapter 20 Heat & 1st Law of Thermo Peer Instruction Time

  4. Physics 1710 — e-Quiz Answer Now ! 0 36% 51 of 140 0 What will happen to a heated ring? The hole will • Get larger • Get smaller • Stay the same

  5. 0 Physics 1710 Chapter 20 Heat & 1st Law of Thermo Time for Real Physics!

  6. 0 Physics 1710 Chapter 20 Heat & 1st Law of Thermo ΔL ΔL Why did it happen? ΔL = α LΔT

  7. 0 Physics 1710Chapter 19 Temperature Ideal Gas Law Idealizations: no interaction between atoms no volume occupied by atoms Number of atoms n = m/M, m = mass; M = molar mass PV = n R T R = 8.315 J/ mol K = 0.08214 L‧atm/mol K = 22.4 L‧atm /273.16 mol K

  8. 0 Physics 1710Chapter 19 Temperature Boltzmann’ Constant k = R/NA k = 1.38 x 10 -23 J/K = 1 yJ/ 7.25 K = 1 eV / 11,600 K PV = N kT

  9. 0 Physics 1710Chapter 19 Temperature Summary: Temperature is a measure of the average kinetic energy of a system of particles. Thermal Equilibrium means that two bodies are at the same temperature. The “Zeroth Law of Thermodynamics” states that if system A and B are n thermal equilibrium with system C, then A and B are in thermal Equilibrium with each other.

  10. 0 Physics 1710Chapter 19 Temperature Kelvin is a unit of temperature where one degree K is 1/279.16 of the temperature of the triple point of water (near freezing). TC = (100/180) (TF – 32 ⁰F) TF = (180/100) TC + 32 ⁰F ∆L/L = α∆T PV = n R T = N kT

  11. 0 Physics 1710 Chapter 20 Heat & 1st Law of Thermo 1′ Lecture: The internal energy is the total average energy of the atoms of an object, average kinetic plus average potential. Heat is the change in internal energy. The change in temperature is proportional to the change in internal energy (heat flow) when there is no change of phase and the system does no work. The first law of thermodynamics states ∆E = ∆Q - W

  12. 0 Physics 1710 Chapter 20 Heat & 1st Law of Thermo 1′ Lecture: Conduction is the flow of kinetic energy from atom to atom. Convection is the transport of energy by bulk motion of atoms. Radiation is the transfer of energy by electromagnetic waves.

  13. 0 Physics 1710 Chapter 20 Heat & 1st Law of Thermo Heat Heat is the macroscopic manifestation of microscopic internal energy. Heat ∆Q in calories (or BTU) 1 calorie (cal) is the amount of energy required to raise the temperature of 0.001 kg of water from 14.5 C to 15.5 C (∆T = 1.00 C). 1 BTU is the heat to raise 1 Lb by 1 ⁰F. James Prescott Joule (1818-1889) showed 1 calorie of heat = 4.186 Joule 1 J = 1 N‧m

  14. 0 Physics 1710 Chapter 20 Heat & 1st Law of Thermo 1 Calorie = 1000 calorie = 1 kcal

  15. 0 Physics 1710 Chapter 20 Heat & 1st Law of Thermo Heat Capacity ∆Q = C ∆T C ≡ dQ/dT Specific Heat c = C /m C ≡ (1/m)dQ/dT ∆Q = (mc)∆T

  16. Physics 1710 — e-Quiz Answer Now ! 0 40% 56 of 140 0 How much heat is required to raise the temperature of 1 kg of water (1 liter) from 20 C to 100 C? Recall c = 1.00 kcal/kgC? • 1.0 kcal • 20 kcal • 80 kcal • 100 kcal • None of the above.

  17. 0 Physics 1710 Chapter 20 Heat & 1st Law of Thermo Guided practice: ∆Q = (mc)∆T How much heat is required to raise the temperature of 1 kg of water (1 liter) from 20 C to 100 C? Recall c = 1.00 kcal/kgC? ∆Q = (mc)∆T ∆Q = (1.00 kg)(1.00 kcal/kg C) (100. C- 20. C) ∆Q = 80. kcal = 80.kcal‧ 4.186 J/cal = 396. kJ

  18. 0 Physics 1710 Chapter 20 Heat & 1st Law of Thermo Change of Phase and Latent Heat It requires an energy “investment” to change the phase from solid to liquid to gas. By breaking the bonds that hold atoms, they can have the same kinetic energy but different total energies. The energy to change the phase is “hidden” and therefore called “latent heat.”

  19. 0 Physics 1710 Chapter 20 Heat & 1st Law of Thermo Latent heat ∆Q = mL For Water: Fusion (and melting) Lf = 333 kJ/kg = 79.4 kcal/kg Vaporization Lv = 2260 kJ/kg = 540 kcal/kg

  20. 0 Physics 1710 Chapter 20 Heat & 1st Law of Thermo Work from a Heat Reservoir The work done by a system is equal to the loss of the internal energy. For an ideal gas W = ∫V1V2 PdV

  21. 0 Physics 1710 Chapter 20 Heat & 1st Law of Thermo First Law of Thermodynamics ∆E = ∆Q -W

  22. 0 Physics 1710 Chapter 20 Heat & 1st Law of Thermo Applications:Adiabatic (∆Q =0) ∆E = -W Isovolumetric (∆V =0) ∆E = ∆Q Isothermal (∆T =0) W =∫V1V2 PdV = ∫V iV f (nRT/V) dVW = n R T ln(Vf /Vi )

  23. 0 Physics 1710 Chapter 20 Heat & 1st Law of Thermo Mid Chapter Summary The internal energy is the total average energy of the atoms of an object. Heat is the change in internal energy. The change in temperature is proportional to the change in internal energy (heat flow) when there is no change of phase and the system does no work. The first law of thermodynamics states ∆E = ∆Q - W

  24. 0 Physics 1710 Chapter 20 Heat & 1st Law of Thermo Conduction: P = kA |dT/dx | Examples: Thermos bottles Blankets Double pane windows Newton’s law of cooling P= h A(T 2 – T1) Pans R factor or R value P= A(T 2 – T1)/∑i Ri

  25. Physics 1710 Chapter 20 Heat & 1st Law of Thermo Convection: Heat transfer by material transfer Forced convection (fluids) External force produces material transfer Natural Convection Buoyancy-driven flow Newton’s law of cooling applied P = h A(T 2 – T1) h depends on flow conditions

  26. Physics 1710 Chapter 20 Heat & 1st Law of Thermo Radiation: Stefan-Boltzmann Law P = εσAT4 Wien’s Law P∝T4 σ = 5.6696 x 10-8 W/m2‧K4 Emissivity 0< ε <1; ε ~ ½ Reflectivity (albedo) R = (1- ε) Energy balance P in - εσA(Tave )4 = 0

  27. Physics 1710 Chapter 20 Heat & 1st Law of Thermo Global Warming?: P in = ( 1- ελ) Psun Tave = [( 1- ελ) Psun /(εGH σA)]1/4 Must understand every parameter to be accurate.

  28. Physics 1710 Chapter 20 Heat & 1st Law of Thermo Summary: Heat is transferred by Conduction—energy diffusion Convection—mass transport Radiation—electromagnetic waves

  29. Physics 1710 Chapter 20 Heat & 1st Law of Thermo P in= ∫0 2π∫0 π/2 I cos θ sin θ r 2dθ dφ P in= I (2πr 2) (½ ) sin2 θ |0 π/2 = I (πr 2) = 1.0 kW/m2 (3.14)(6.4x106 m)2 = 1.3 x 1017 W P in = εσA(Tave )4 Tave = [P in/ εσA]1/4 = [1.3 x 1017 /(0.6‧5.66 x10-8‧ 5.1x1014)] 1/4 =294 K=21C Guided Practice If the power per unit area (intensity) of sunlight on the earth is 1.0 kW/m2 and the emissivity is 0.6, what is the expected avearge temperatue of the earth? Comment on the effect of a change in the emissivity or solar radiation.

  30. Physics 1710 Chapter 20 Heat & 1st Law of Thermo Guided Practice If the power per unit area (intensity) of sunlight on the earth is 1.0 kW/m2 and the emissivity is 0.6, what is the expected average temperature of the earth? Comment on the effect of a change in the emissivity or solar radiation.

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