2. Chapter 10-Thermodynamics

3.  Describe the relationship between heat, work, and internal energy  Make calculations involving work, pressure, and volume for a gas Learning Targets 10.1 Relationships between Heat and Work

4. Background "thermo": Greek therme heat "dynamics": Greek dynamikos powerful Physics that deals with the mechanical action or relations between heat and work Example : Heat to work Heat Q from flame provides energy to do work 10.1 Relationships between Heat and Work

5. THERMODYNAMICS: the science of energy, specifically heat and work, and how the transfer of energy effects the properties of materials. A “system” is the “collection of objects on which attention is being focused” The “surroundings” are everything else in the environment The system and surroundings must be separated by walls which can either insulate or allow heat flow

6. Heat... is the amount of internal energy entering or leaving a system... occurs by conduction, convection, or radiation.... causes a substance's temperature to change... is not the same as the internal energy of a substance... is positive if thermal energy flows into the substance... is negative if thermal energy flows out of the substance... is measured in joules 10.1 Relationships between Heat and Work

7. Thermal Equilibrium SS ystems (or objects) are said to be in thermal equilibrium if there is no net flow of thermal energy from one to the other. AA thermometer is in thermal equilibrium with the medium whose temperature it measures, for example. II f two objects are in thermal equilibrium, they are at the same temperature. 10.1 Relationships between Heat and Work

8. Work, W, energy caused by physical motion 10.1 Relationships between Heat and Work

9. WORK W is positive if work is done by system. Air does work on the environment: W > 0. W is negative if work is done on the system. Environment (man) does work on system: W < 0 10.1 Relationships between Heat and Work

10.  The terms heat and work always refer to energy in transit o An object never possesses “heat” or “work”; it only has internal energy  Energy is always added to or taken away from the system by its surroundings, unless both objects have the same temperature Heat and Work 10.1 Relationships between Heat and Work

11.  Why does pulling a nail from a piece of wood cause the temperature of the nail and the wood to increase? 10.1 Relationships between Heat and Work Example

12.  When a nail is pulled from a piece of wood, work is done by the frictional forces between the nail and the wood fibers  This work increases the internal energy of the iron atoms in the nail and the molecules in the wood, thus increasing their temperature  As a result, energy is transferred as heat from the nail to the surrounding air until both are at the same temperature 10.1 Relationships between Heat and Work

13.  As illustrated by the previous example, work can increase the internal energy of a substance  This internal energy can then decrease through the transfer of energy as heat to another substance.  This energy can then be used to do work Changes in Internal Energy 10.1 Relationships between Heat and Work

14.  System : flask+ balloon+ water  The burner transferred energy as heat to the system, which cause the internal energy to increase  When the water boils and becomes steam, the steam’s volume increases and the balloon does work on the atmosphere  The internal energy of the system decreased o Some of the energy transferred as heat was transferred out of the system as work done on the air. 10.1 Relationships between Heat and Work Example : Heating a flask of water with a balloon on top

15. Pressure is a measure of how much force is applied over a given area Volume is how much space something takes up. The work done on or by a gas is calculated as: Work = Pressure x volume change W = P∆V This definition of work assumes that pressure remains constant and only volume changes Work, Pressure, and Volume 10.1 Relationships between Heat and Work

16.  If the gas expands, ∆V is positive, and work is done by the gas  If the gas is compressed, ∆V is negative, and the work is done on the gas  When the gas volume remains constant, there is no displacement and no work is done on or by the system 10.1 Relationships between Heat and Work

17.  Although pressure can change during a process, work is done only if the volume changes (displacement)  A situation in which pressure increases and volume remains constant is comparable to one in which a force does not cause a displacement 10.1 Relationships between Heat and Work

18. Work done by a movable piston Work=F x d = F/A x d x A = P DV 10.1 Relationships between Heat and Work

19.  An engine cylinder has a volume of 0.0004 m 3. How much work can be done by a gas in a cylinder if the gas exerts a constant pressure of 7.5 x 10 5 N/m 2 on the piston? Sample Problem 10.1 Relationships between Heat and Work

20. Thermodynamics Processes 10.1 Relationships between Heat and Work Bomb Calorimeter: -this process takes place inside this device, which is a thick container in which a small quantity of a substance undergoes a combustion reaction. -When energy is released by the reaction, it increases the pressure and temperature of the gaseous reaction. Because of the thick walls in the container, there is no change in volume of gas. -Energy can be transferred to or from the container only as heat. ISOVOLUMETRIC PROCESS Interpretation Energy added to the system as heat (Q>0) increases the system’s internal energy Energy removed from the system as heat (Q<0) decreases the system’s internal energy. By: Hedieh Anna Brittany Timo

21. Thermodynamics Processes 10.1 Relationships between Heat and Work An isochoric process is one where the volume of the system stays constant. Again, 'iso' means the same and 'choric' means volume. Volume is the amount of space the material takes up. So this would be like heating a gas in a solid, non-expandable container. The molecules would move faster and the pressure would increase, but the size of the container stays the same.

22.  Isothermal Process  An isothermal process is one where the temperature of the system stays constant. Thermal relates to heat, which is in turn related to temperature. Temperature is the average heat (movement) energy of the molecules in a substance.  An example of an isothermal process would be if we took a gas held behind a movable piston and compressed that piston: the volume has decreased, and the pressure behind the piston has increased, since the molecules have less space in which to move. When you compress a piston, you're using energy - you're doing work on the gas - so normally the molecules would gain energy and move faster, and the temperature would increase. So the only way for an isothermal process to happen is if all that energy you put into compressing the gas comes out again, for example by putting a cold reservoir in contact with the piston.

23. ISOCHORIC EXAMPLE: Heat input increases P with const. V 400 J heat input increases internal energy by 400 J and zero work is done. 400 J No Change in volume: 10.1 Relationships between Heat and Work

25. The First Law of Thermodynamics states that : The internal energy of a system changes from aninitial value U i to a final value U f due to heat added ( Q) and work done by the system ( W)  U = U f – U i = Q – W QQ is positive when the system gains heat, and negative when the system loses heat. WW is positive when it is done BY the system, and negative when it is done ON the system

26. Example: 1000 J of thermal energy flows into a system (Q = 1000 J). At the same time, 400 J of work is done by the system (W = 400 J). What is the change in the system's internal energy U? ---------------------------------------------------------- Solution:  U = Q - W = 1000 J - 400 J = 600 J

27. Example: 800 J of work is done on a system (W = -800 J) as 500 J of thermal energy is removed from the system (Q = -500 J). What is the change in the system's internal energy U? ----------------------------------------------------- Solution:  U = Q - W = -500 J - (-800 J) = -500 J + 800 J = 300 J

28. Work Done by an Expanding Gas W = P  V  V = V f - V i W = P (V f - V i ) Area under pressure-volume curve is the work done ----------------------------------------- Isobaric Process: "same pressure" Greek: barys, heavy

29. Work and the Pressure-Volume Curve Work Done = Area Under PV curve ------------------------------------- How much work is done by the system when the system is taken from: (a) A to B (900 J) (b) B to C (0 J) (c) C to A (-1500 J) ------------------------------------- Each "rectangle" has an area of 100 Pa-m 3 = 100 (N/m 2 )-m 3 = 100 N-m = 100 Joules

30. Expanding Gas Example: If a gas expands at a constant pressure, the work done by the gas is: W = P  V 10 grams of steam at 100 o C at constant pressure rises to 110 o C: P = 4 x 10 5 Pa     T = 10 o C  V = 30.0 x 10 -6 m 3 c = 2.01 J/g o C What is the change in internal energy?  U = Q - W W = (4 x 10 5 )(30.0 x 10 -6 ) = 12 J Q = mc  T = (10)(2.01)(10) = 201 J  U = Q - W = 201 J - 12 J = 189 J

31. Work, Rubber Bands, and Internal Energy  U = Q - W Expand rubber band: W 0 temperature increases ------------------------------------------- Press thick rubber band to forehead and expand it rapidly. The warming should be obvious. Now allow the band to contract quickly; cooling will also be evident.

33. ISOTHERMAL-Temperature remains constant

34. ISOBARIC - Pressure remains constant

35. ISOMETRIC - Volume remains constant (also ISOVOLUMETRIC or ISOCHORIC) Since ΔV = 0, W = 0 then  U = Q - W = Q

36. Adiabatic Expansion of a Ideal Gas No heat transfer therefore no temperature change (Q=0). Generally obtained by surrounding the entire system with a strongly insulating material or by carrying out the process so quickly that there is no time for a significant heat transfer to take place. If Q = 0 then ΔU = - W A system that expands under adiabatic conditions does positive work, so the internal energy decreases. A system that contracts under adiabatic conditions does negative work, so the internal energy increases.

37. Adiabatic Expansion of a Ideal Gas Blowing air through wide open mouth results to warm air. Blowing through small opening results to cooler air due to adiabatic expansion. Both adiabatic expansion and compression of gases occur in only hundredths of a second in the cylinders of a car’s engine. Compresses air leaking out through a small opening also results in adiabatic cooling.

38. PROCESS DIAGRAMS: visualize processes using properties (T, P, V, etc.) Area underneath the slope represents the amount of work done (P x V).

39. CYCLE: a system undergoes processes - returning to its initial state Area underneath the slope represents the amount of work done (P x V).

41. RRefrigerators work by taking heat from the interior and depositing it on the exterior TThe compressor raises the pressure and temperature of the refrigerant (freon or ammonia) while the coils OUTSIDE the refrigerator allow the now hot refrigerant to dissipate the heat TThe warm refrigerant flows through an expansion valve from a high-pressure to a low- pressure zone, so it expands and evaporates T The coils INSIDE the refrigerator allow the cold refrigerant to absorb heat, cooling the interior he cool refrigerant flows back to the compressor, and the cycle repeats

42. Second Law of Thermodynamics Heat flows naturally from a region at high temperature to a region at low temperature. By itself, heat will not flow from a cold to a hot body. When an isolated system undergoes a change, passing from one state to another, it will do so in such a way that its entropy (disorder) will increase, or at best remain the same.

43. ENTROPY

44. Can you beat the Second Law? So, can you cool your kitchen by leaving the refrigerator door open So, can you cool your kitchen by leaving the refrigerator door open NO! NO! The heat removed from the interior of the refrigerator is deposited back into the kitchen by the coils on the back! The heat removed from the interior of the refrigerator is deposited back into the kitchen by the coils on the back! And to make matters worse, the Second Law of Thermodynamics says that work is needed to move the heat from cold to hot, so the actual amount of heat added to the kitchen is MORE than the amount removed from the refrigerator And to make matters worse, the Second Law of Thermodynamics says that work is needed to move the heat from cold to hot, so the actual amount of heat added to the kitchen is MORE than the amount removed from the refrigerator

46. Hopefully, you understand today’s lesson. Otherwise, you’ll end up like this cow.