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Work, Heat, and the First Law of Thermodynamics

This chapter discusses the concepts of work, heat, and the first law of thermodynamics in relation to mechanical energy and thermal energy. It covers topics such as conservative forces, friction forces, work done by external forces, thermal interactions, and the relationship between heat, temperature, and thermal energy. Examples and calculations are provided to illustrate the principles of thermodynamics.

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Work, Heat, and the First Law of Thermodynamics

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  1. Chapt. 17. Work, Heat, and the First Law of Thermodynamics • Mechanical Energy Emech = K + U • If there are only conservative forces ( ex. Gravity force, spring force) in the system ΔEmech = ΔK + ΔU = 0 • If there is friction forces ΔEmech = ΔK + ΔU = Wfric < 0 • General case ΔEmech + ΔEth = Wext • ΔEth is the change of thermal energy and Wext is the work done by the external force. • The systems we want to study in thermodynamics are stationary containers gases or liquids, whose center-of –mass mechanical energy does not change. Thus ΔEmech = 0

  2. Stop to think 17.1 page 508Stop to think 17.2 page 513Stop to think 17.3 page 516Stop to think 17.5 page 521Stop to think 17.6 page 529 • Example 17.2 page 512 • Example 17.3 page 520 • Example 17.4 page 521 • Example 17.7 page 524

  3. Thermal Energy • Thermal energy Eth is associated with the system’s temperature. We now need to be a bit more specific • During a phase change, from solid to liquid or from liquid to gas, a system’s thermal energy increases but its temperature does not. • If there are no phase changes, increasing the system’s temperature increase its thermal energy. • A system’s thermal energy does not change (ΔEth = 0) during an isothermal process (ΔT =0)

  4. 0.10 mol of a montomic gas follows the process shown in the Figure, What is the total change in thermal energy of the gas?

  5. Work • Work is the energy transferred between a system and environment when a net force acts on the system over a distance. • The sign of the work • Work is positive when the force is in the direction of motion • Work is negative when the force is opposite to the motion

  6. Heat • The energy transferred in a thermal interaction is called heat • The symbol for heat is Q • The energy equation now becomes • ΔEsys = ΔEmech +ΔEth = Wext + Q Quick quiz: A gas cylinder and piston are covered with heavy insulation. The piston is pushed into the cylinder, compressing the gas. In the process. The gas temperature • Increases • Decreases • Doesn’t change

  7. Work in Ideal-Gas Processes • The work done on the system • When we press the gas, the gas volume becomes smaller, so the total work done by the environment on the gas

  8. Work in some special processes • Isochoric Process W = 0 • Isobaric Process W = -PΔV • Isothermal Process Work depends on path

  9. Finding work from the P-V diagram • W = the negative of the area under the PV curve between Vi and Vf W < 0 W > 0

  10. Heat and Thermal interactions • Heat is the energy transferred during a thermal interaction • Units of heat • The SI unit of heat is joule. • Historically, unit for measuring heat, is calorie • A cal = the quantity of heat needed to change the temperature of 1 g of water by 1 oC. • 1cal = 4.186 J 1 food calorie = 1 Cal = 1000 cal =1 kcal

  11. Distinguish between heat, temperature, and thermal energy • Thermal energy is an energy of the system due to the motion of its atoms and molecules. Thermal energy is a state variable, it may change during a process. The system’s thermal energy continues to exist even if the system is isolated and not interacting thermally with its environment • Heat is energy transferred between the system and the environment as they interact. Heat is not a particular form of energy, nor is it a state variable. Heat may cause the system’s thermal energy to change, but that does not mean that heat and thermal energy are the same thing. • Temperature is a state variable, it is related to the thermal energy per molecule. But not the same thing.

  12. The first Law of thermodynamics • ΔEsys = ΔEmech +ΔEth = W+ Q • Here we assume the system mechanical energy does not change ΔEmech = 0 • ΔEth = W+ Q = work on system + heat to system ( first law of thermodynamics) • The first law of thermodynamics is the law of conservation of energy. The first law of thermodynamics doesn’t tell us anything about the value of Eth only how Eth changes, doing 1 J of work changes the thermal energy by ΔEth = 1 J. • The system’s thermal energy isn’t the only thing that changes. Work or heat that change the thermal energy also change the pressure, volume, temperature, and other state variables. The first law tells us only about ΔEth . Other law and relationship must be used to learn how the other state variables change.

  13. First-Law bar chart • The isochoric process. • The final point is on a lower isotherm than the initial point, that is Tf < Ti. No work is done (W=0) in this process. Heat energy was transferred out of gas (Q < 0) and thermal energy of gas decreased (ΔE < 0) as the temperature fell. • ΔEth = E (th f) – E( th i) = W + Q • The isothermal process. ΔT = 0 ΔEth = 0 , therefore W = -Q The heat energy is transfered, causes the gas to expand and do the work to environment

  14. Specific Heat What happens to a system when you change its thermal energy? • The temperature of the system changes • The system undergoes a phase change, such as melting or freezing Specific Heat C : The amount of energy that raises the temperature of 1 Kg of A substance by 1 K ( or 1C) is called specific heat. In other words, the thermal energy of the system changes by ΔEth = Mc ΔT ( Kelvin and Celsius temperature scales have the same step size. ) The first law of thermodynamics ΔEth = W+ Q , In working with solids And liquids, we almost always change the temperature by heating, so W = 0, Then the heat needed to bring about a temperature change ΔT is Q = Mc ΔT . Molar specific heat: The amount of energy that raises the temperature of 1 mol of a substance by 1K. It depends on the molar mass. Table 17.2 (page 527) list few specific heat and molar specific heats of solids and Liquids.

  15. Phase change and heat of transformation • A phase change is characterized by a change in thermal energy without change in temperature • Heat of fusion Lf : the heat of transformation for 1 Kg substance between a solid and liquid. • Heat of vaporization Lv the heat of transformation for 1 Kg substance between a liquid and gas. • Q = ± M Lf melt/freeze • Q = ± M Lv boil/condense • Table 17.3 list few melting/boiling temperatures and heat of transformation. • For systems that undergo a temperature change, Q = Mc (Tf – Ti). • For systems that undergo a phase change Q = ± ML, Supply the correct sign by observing whether energy enters or leaves the system during the transition.

  16. Problem 44: your 300 ml cup of coffee is too hot to drink when served at 90 C. what is the mass of an ice cue, taken from a –20C freezer, that will cool your coffee to a pleasant 60 C • There are two interacting systems: coffee (i.e., water) and ice. Changing the coffee temperature from 90°C to 60°C requires four steps: (1) raise the temperature of ice from -20°C to 0°C, (2) change ice at 0°C to water at 0°C, (c) raise the water temperature from 0°C to 60°C, and (4) lower the coffee temperature from 90°C to 60°C. • Solve: For the closed coffee-ice system,Q = Q(ice) + Q(coffee) = 0 • Q (ice) = Q (ice, -20C –0) + Q ( melting) + Q(water, 0 –60c) • Q (coffee) = m(cof)C(water) (60-90) • Q (coffee) = 0.3 Kg (4190 J?kg) ( -30) = -37,710 J • Suppose the mass of ice is M • Q (ice) = 41800 M + 330,000 M + 251,400 M = 37710 J • M = 0.0605 kg = 60.5 g

  17. Quiz: Objects A and B are brought into close thermal contact with each other, but they are well isolated from their surroundings. Initially TA = 0 oC and TB = 10 0 oC. The specific heat of A is less than the specific heat of B. The two objects will soon reach a common final temperature Tf . The final temperature is a.Tf > 50 oC b. Tf = 50 oC c. Tf < 50 oC Answer: a QA + QB = 0 MACA(Tf –0) + MBCB(Tf –100)=0 CA(Tf –0) = CB(100-Tf)

  18. The specific heats of gases • The heat required to cause a specified temperature change depends on the process by which the gas changes Process A and B, which start on the Ti And end on the Tf, have the same temperature Change ÄT = Tf-Ti, But different amount of Heat is required. ÄT at constant volume ÄT at constant pressure Cv is the molar specific heat at constant Volume. Cpis the molar specific heat at constant pressure

  19. Cp and Cv • ΔEth (constant volume process) = W+ Q = 0 + nCvΔT = nCvΔT • ΔEth (constant pressure process) = W+ Q = -pΔV + nCpΔT • If both have the same ΔT, ΔEth in tow process is the same • -pΔV + nCpΔT = nCvΔT • Using the ideal gas law, in constant-pressure process pΔV = nR ΔT -nR ΔT + nCpΔT = nCvΔT Cp = Cv + R ΔEth = nCvΔT (any idea-gas process)

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