1 / 26

What is the minimum number of additional dietary Calories (kcal) that a 100 kg mountain climber must burn to scale a 100

Physics 1710 —Warm-up Quiz Answer Now ! 0 34% 48 of 140 0 What is the minimum number of additional dietary Calories (kcal) that a 100 kg mountain climber must burn to scale a 1000 m cliff? About 29 Cal About 230 Cal About 1000 Cal About 29,000 Cal About 230,000 Cal 0

Audrey
Télécharger la présentation

What is the minimum number of additional dietary Calories (kcal) that a 100 kg mountain climber must burn to scale a 100

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Physics 1710—Warm-up Quiz Answer Now ! 0 34% 48 of 140 0 What is the minimum number of additional dietary Calories (kcal) that a 100 kg mountain climber must burn to scale a 1000 m cliff? • About 29 Cal • About 230 Cal • About 1000 Cal • About 29,000 Cal • About 230,000 Cal

  2. 0 Physics 1710Chapter 21 Kinetic theory of Gases Solution: ∆E = ∆Q – W = 0 (no change in climbers internal energy) So: ∆Q = W (in this case) W = mgh W = 100 kg (9.80 N/kg)(1000 m) = 9.80 x 10 5 J ∆Q = (9.80 x 10 2 kJ)/(4.183 kJ/kcal) ∆Q = 234. kcal

  3. 0 Physics 1710 Chapter 20 Heat & 1st Law of Thermo • Examples: • Touching a hot stove • Feeling the air rising from it • Feeling the glow Conduction Convection Radiation How does heat get from one place to another?

  4. 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

  5. 0 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

  6. 0 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

  7. 0 Physics 1710 Chapter 20 Heat & 1st Law of Thermo Summary: Heat is transferred by Conduction—energy diffusion Convection—mass transport Radiation—electromagnetic waves

  8. No Talking! Think! Confer! 0 How does heat get to the earth from the sun? What factors are important in the average temperature of the planet? (1) conduction; (2) convection; (3) radiation; (4) conduction and convection; (5) convection and radiation. Physics 1710Chapter 21 Kinetic theory of Gases Peer Instruction Time

  9. Physics 1710 — e-Quiz Answer Now ! 10 35% 50 of 140 0 How does heat get to the earth from the sun? What factors are important in the average temperature of the planet? • Conduction • Convection • Radiation • Conduction and convection • Convection and Radiation

  10. 0 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.

  11. 0 Physics 1710Chapter 21 Kinetic theory of Gases 1′ Lecture: The Ideal Gas Law results from the cumulative action of atoms or molecules. The average kinetic energy of the atoms or molecules of an ideal gas is equal to 3/2 kT. Energy average distributes equally (is equipartitioned) into all available states. The distribution of particles among available energy states obeys the Boltzmann distribution law. nV = no e –E/kT

  12. 0 Physics 1710Chapter 21 Kinetic theory of Gases Molecular Model of Ideal Gas Key concept: gas is ensemble of non-interacting atoms or molecules. Pressure due to a single molecule at wall of vessel: P1 = -F1 /A = -(∆px / ∆t)/A Impulse: ∆px = - mvx – (mvx ) = -2 mvx ∆t = 2d /vx Thus: P1 = - F1 /A = mvx2 /(d‧A)

  13. 0 Physics 1710Chapter 21 Kinetic theory of Gases P1 = - F1 /A = mvx2 /(d‧A) Total Pressure: P = N<P1 >= Nm<vx2 >/(d‧A) P = (N/V) m<vx2 > Average vx2 = <vx2 >: <v2 >=<vx2 > + <vy2 > + <vz2 > <v2 >=3<vx2 >; <vx2 >= 1/3 <v 2> P = ⅔(N/V)(½ m<v 2>)

  14. 0 Physics 1710Chapter 21 Kinetic theory of Gases P = ⅔(N/V)(½ m<v2 >) But P = (N/V) kT Thus T = 2/(3k)(½ m<v 2>) ½ m<v2> = 3/2 kT ½ m<vx2> = 1/3 [½ m<v2> ] = ½ kT

  15. 0 Physics 1710Chapter 21 Kinetic theory of Gases Principle of Equipartition of Energy: Each degree of freedom contributes 1/2 kT to the energy of a system.

  16. 0 Physics 1710Chapter 21 Kinetic theory of Gases Each molecule in a gas contributes 3 degrees of freedom to the system: N( 1/2 m<v 2>) = 3N(½ kT) = 3/2 nRT √<v 2> = vrms = √(3 kT/m) = √(3RT/M)

  17. 0 Physics 1710Chapter 21 Kinetic theory of Gases (Molar) Specific Heat of an Ideal Gas: ∆Q = n CV ∆T (constant volume) ∆Q = n CP ∆T (constant pressure) W = ∫ P dV; at constant volume W = 0. ∆Eint = ∆Q = n CV ∆T Eint = n CV T CV = (1/n) d Eint /dT CV = 3/2 R = 3/2 No kT = 12.5 J/mol‧K

  18. 0 Physics 1710Chapter 21 Kinetic theory of Gases ∆Eint = ∆Q –W = nCP ∆T - P∆V nCV ∆T= nCP ∆T – n R ∆T CP - CV = R CP = 5/2 R γ = CP / CV = (5/2 R)/(3/2 R) = 5/3 γ = 5/3

  19. 0 Physics 1710Chapter 21 Kinetic theory of Gases Adiabatic Expansion of an Ideal Gas: For adiabatic case: dEint = n CV dT = - PdV So that dT = -P dV /(nCV ) Also: PV = nRT PdV + VdP = nR dT PdV + VdP = -RP /(nCV ) dV

  20. 0 Physics 1710Chapter 21 Kinetic theory of Gases PdV + VdP = -R/(nCV ) PdV Rearranging: dP/P = [1 – R/(nCV)] dV/V dP/P = - γ dV/V ln P = - γ lnV + ln K PV γ= constant

  21. 0 Physics 1710Chapter 21 Kinetic theory of Gases Bulk Modulus of an Ideal Gas: B = -∆P/ (∆V/V) P= K V - γ dP = - γ KV - γ - 1dV B = -dP/(dV/V) B= (γ KV 1dV)/(dV/V) B= γ KV - γ B = γ P

  22. 0 Physics 1710Chapter 21 Kinetic theory of Gases Velocity of Sound in Ideal Gas: v = √(B/ρ) v = √[ γP/(ρo P/ Po )] v = √[ γPo /ρo ]N.B.: no pressure dependence For air (ideal) v = √[ γPo /ρo ]v = √[(5/3)(101 kPa/ 1.26 kg/m3 )]v = 365 m/s (Cf 343 m/s)

  23. 0 Physics 1710Chapter 21 Kinetic theory of Gases Law of Atmospheres dP = -mg nV dy P = nV kT dP = kT dnV kT dnV= -mg nV dy dnV/ nV = -(mg/kT) dy nV = no e –(mgy/kT)

  24. 0 Physics 1710Chapter 21 Kinetic theory of Gases Boltzmann Distribution Function nV = no e –(mgy/kT) nV = no e –U/kT nV (E) = no e –E/kT

  25. 0 Physics 1710Chapter 21 Kinetic theory of Gases Summary: The Ideal Gas Law results from the cumulative action of atoms or molecules. The average kinetic energy of the atoms or molecules of an ideal gas is equal to 3/2 kT. ½ m<v2> = 3/2 kT Energy average distributes equally (is equipartitioned) into all available states. Each degree of freedom contributes 1/2 kT to the energy of a system.

  26. 0 Physics 1710Chapter 21 Kinetic theory of Gases Summary (cont’d.) γ = CP / CV PV γ= constant B = γ P The distribution of particles among available energy states obeys the Boltzmann distribution law. nV = no e –E/kT

More Related