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

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