# Lecture 24. Paths on the pv diagram

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1 Goals: Lecture 24 Chapter 17 Apply heat and energy transfer processes Recognize adiabatic processes Chapter 18 Follow the connection between temperature, thermal energy, and the average translational kinetic energy molecules Understand the molecular basis for pressure and the idealgas law. To predict the molar specific heats of gases and solids. Assignment HW10, Due Wednesday 9:00 AM For Thursday, Read through all of Chapter 18 Physics 207: Lecture 24, Pg 1 Paths on the p diagram (4) Isobaric (3) Isothermal (2) Isochoric (1) Adiabatic W = - p???? W = 0???? p T 1 T 2 Ideal gas T 4 T 3 Physics 207: Lecture 24, Pg 2 Page 1

2 Isothermal processes Work done when P = nrt = constant P = nrt / W W = = final initial f i p d nrt = (area under curve) d / = nrt f i d / W = nrt ln( / ) f i p 3 T 1 T 2 T 3 T 4 Physics 207: Lecture 24, Pg 3 Adiabatic Processes An adiabatic process is process in which there is no thermal energy transfer to or from a system (Q = 0) A reversible adiabatic process involves a worked expansion in which we can return all of the energy transferred. In this case P γ = const. All real processes are not. p T 1 T 2 T 4 T 3 Physics 207: Lecture 24, Pg 4 Page 2

3 Work and Ideal Gas Processes (on system) Isothermal W = nrt ln( / ) f i Isobaric W Isochoric W = p = 0 ( f - ) FYI: Adiabatic (and reversible) W 2 = Pd = 1 i 2 1 const γ d = const γ γ γ ( 2 1 Physics 207: Lecture 24, Pg 5 ) Combinations of Isothermal & Adiabatic Processes All engines employ a thermodynamic cycle W = ± (area under each p curve) W cycle = area shaded in turquoise Watch sign of the work! Physics 207: Lecture 24, Pg 6 Page 3

4 Relationship between energy transfer and T Physics 207: Lecture 24, Pg 7 Heat and Latent Heat Latent heat of transformation L is the energy required for 1 kg of substance to undergo a phase change. (J / kg) Q = ±ML Specific heat c of a substance is the energy required to raise the temperature of 1 kg by 1 K. (Units: J / K kg ) Q = M c T Molar specific heat C of a gas at constant volume is the energy required to raise the temperature of 1 mol by 1 K. Q = n C T If a phase transition involved then the heat transferred is Q = ±ML+M c T Physics 207: Lecture 24, Pg 8 Page 4

5 Q : Latent heat and specific heat The molar specific heat of gasses depends on the process path C = molar specific heat at constant volume C p = molar specific heat at constant pressure C p = C +R (R is the universal gas constant) γ = C p C Physics 207: Lecture 24, Pg 9 Mechanical equivalent of heat Heating liquid water: Q = amount of heat that must be supplied to raise the temperature by an amount T. [Q] = Joules or calories. 1 Cal = J 1 kcal = 1 Cal = 4186 J calorie: energy to raise 1 g of water from 14.5 to 15.5 C (James Prescott Joule found the mechanical equivalent of heat.) Sign convention: +Q : heat gained - Q : heat lost Physics 207: Lecture 24, Pg 10 Page 5

6 Exercise The specific heat (Q = M c T) of aluminum is about twice that of iron. Consider two blocks of equal mass, one made of aluminum and the other one made of iron, initially in thermal equilibrium. Heat is added to each block at the same constant rate until it reaches a temperature of 500 K. Which of the following statements is true? (a) The iron takes less time than the aluminum to reach 500 K (b) The aluminum takes less time than the iron to reach 500 K (c) The two blocks take the same amount of time to reach 500 K Physics 207: Lecture 24, Pg 11 Heat and Ideal Gas Processes (on system) Isothermal Expansion/Contraction E Th = 0 = W + Q Q = W Isobaric Q = nc T = n( C + R) T p Isochoric Q = nc T Adiabatic Q = 0 Physics 207: Lecture 24, Pg 12 Page 6

7 Two process are shown that take an ideal gas from state 1 to state 3. Compare the work done by process A to the work done by process B. A. W A > W B B. W A < W B C. W A = W B = 0 D. W A = W B but neither is zero ON BY A 1 3 W 1 2 = 0 (isochoric) B 1 2 W 1 2 = -½ (p 1 +p 2 )( 2-1 ) < 0 -W 1 2 > 0 B 2 3 W 2 3 = -½ (p 2 +p 3 )( 1-2 ) > 0 -W 2 3 < 0 B 1 3 = ½ (p 3 - p 1 )( 2-1 ) > 0 < 0 Physics 207: Lecture 24, Pg 13 Exercise Latent Heat Most people were at least once burned by hot water or steam. Assume that water and steam, initially at 100 C, are cooled down to skin temperature, 37 C, when they come in contact w ith your skin. Assume that the steam condenses extremely fast, and that the specific heat c = 4190 J/ kg K is constant for both liquid water and steam. Under these conditions, which of the following statements is true? (a) Steam burns the skin worse than hot water because the thermal conductivity of steam is much higher than that of liquid water. (b) Steam burns the skin worse than hot water because the latent heat of vaporization is released as well. (c) Hot water burns the skin worse than steam because the thermal conductivity of hot water is much higher than that of steam. (d) Hot water and steam both burn skin about equally badly. Physics 207: Lecture 24, Pg 14 Page 7

8 Energy transfer mechanisms Thermal conduction (or conduction) Convection Thermal Radiation For a material of cross-section area A and length L, spanning a temperature difference T = T H T C, the rate of heat transfer is Q / t = k A T / x where k is the thermal conductivity, which characterizes whether the material is a good conductor of heat or a poor conductor. Physics 207: Lecture 24, Pg 15 Energy transfer mechanisms Thermal conduction (or conduction): Energy transferred by direct contact. e.g.: energy enters the water through the bottom of the pan by thermal conduction. Important: home insulation, etc. Rate of energy transfer ( J / s or W ) Through a slab of area A and thickness x, with opposite faces at different temperatures, T c and T h Q / t = k A (T h - T c ) / x k :Thermal conductivity (J / s m C) Physics 207: Lecture 24, Pg 16 Page 8

9 Thermal Conductivities Aluminum J/s m C J/s m C J/s m C 238 Air Asbestos 0.25 Copper 397 Helium Concrete 1.3 Gold 314 Hydrogen Glass 0.84 Iron 79.5 Nitrogen Ice 1.6 Lead 34.7 Oxygen Water 0.60 Silver 427 Rubber 0.2 Wood 0.10 Physics 207: Lecture 24, Pg 17 Home Exercise Thermal Conduction Two identically shaped bars (one blue and one green) are placed between two different thermal reservoirs. The thermal conductivity coefficient k is twice as large for the blue as the green. You measure the temperature at the joint between the green and blue bars. 100 C T joint 300 C Which of the following is true? (A) T top > T bottom (B) T top = T bottom (C) T top < T bottom (D) need to know k Physics 207: Lecture 24, Pg 18 Page 9

10 Home Exercise Thermal Conduction Two identically shaped bars (one blue and one green) are placed between two different thermal reservoirs. The thermal conductivity coefficient k is twice as large for the blue as the green. 100 C T joint 300 C Top: P green = P blue = Q / t = 2 k A (T high - T j ) / x= k A (T j - T low ) / x 2 (T high - T j ) = (T j - T low ) 3 T j(top) = 2 T high T low By analogy for the bottom: 3 T j(bottom) = 2 T low T high 3 (T j(top) - T j(bottom ) = 3 T high 3 T low > 0 (A) T top > T bottom Physics 207: Lecture 24, Pg 19 Exercise Thermal Conduction Two thermal conductors (possibly inhomogeneous) are butted together and in contact with two thermal reservoirs held at the temperatures shown. Which of the temperature vs. position plots below is most physical? 100 C (A) (B) (C) 300 C Temperature Position Temperature Position Temperature Position Physics 207: Lecture 24, Pg 20 Page 10

11 Energy transfer mechanisms Convection: Energy is transferred by flow of substance 1. Heating a room (air convection) 2. Warming of North Altantic by warm waters from the equatorial regions Natural convection: from differences in density Forced convection: from pump of fan Radiation: Energy is transferred by photons e.g.: infrared lamps Stefan s Law P = σ A e T 4 (power radiated) σ = W/m 2 K 4, T is in Kelvin, and A is the surface area e is a constant called the emissivity Physics 207: Lecture 24, Pg 21 Minimizing Energy Transfer The Thermos bottle, also called a Dewar flask is designed to minimize energy transfer by conduction, convection, and radiation. The standard flask is a double-walled Pyrex glass with silvered walls and the space between the walls is evacuated. acuum Silvered surfaces Hot or cold liquid Physics 207: Lecture 24, Pg 22 Page 11

12 Anti-global warming or the nuclear winter scenario Assume P/A = I = 1340 W/m 2 from the sun is incident on a thick dust cloud above the Earth and this energy is absorbed, equilibrated and then reradiated towards space where the Earth s surface is in thermal equilibrium with cloud. Let e (the emissivity) be unity for all wavelengths of light. What is the Earth s temperature? P = σ A T 4 = σ (4π r 2 ) T 4 = I π r 2 T = [I / (4 x σ )] ¼ σ = W/m 2 K 4 T = 277 K (A little on the chilly side.) Physics 207: Lecture 24, Pg 23 Ch. 18, Macro-micro connection Molecular Speeds and Collisions A real gas consists of a vast number of molecules, each moving randomly and undergoing millions of collisions every second. Despite the apparent chaos, averages, such as the average number of molecules in the speed range 600 to 700 m/s, have precise, predictable values. The micro/macro connection is built on the idea that the macroscopic properties of a system, such as temperature or pressure, are related to the average behavior of the atoms and molecules. Physics 207: Lecture 24, Pg 24 Page 12

13 Molecular Speeds and Collisions A view of a Fermi chopper Physics 207: Lecture 24, Pg 25 Molecular Speeds and Collisions Physics 207: Lecture 24, Pg 26 Page 13

14 Mean Free Path If a molecule has N coll collisions as it travels distance L, the average distance between collisions, which is called the mean free path (lowercase Greek lambda), is Physics 207: Lecture 24, Pg 27 Macro-micro connection Assumptions for ideal gas: # of molecules N is large They obey Newton s laws Short-range interactions with elastic collisions Elastic collisions with walls (an impulse..pressure) What we call temperature T is a direct measure of the average translational kinetic energy What we call pressure p is a direct measure of the number density of molecules, and how fast they are moving (v rms ) v 2 T = ε 3 rms k B p = = 2 3 avg N ε 2 ( v ) avg avg = 3k BT m Physics 207: Lecture 24, Pg 28 Page 14

15 Lecture 24 Assignment HW10, Due Wednesday (9:00 AM) Tuesday review Reading assignment through all of Chapter 18 Physics 207: Lecture 24, Pg 29 Page 15

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