A heat pump takes in 440J of heat from a low temp reservoir in each cycle and uses 200J of work to move the heat to a higher temp reservoir. How much heat is released to the higher temp reservoir in each cycle in J?
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- Consider these scenarios and state whether work is done by the system on the environment (SE) or by the environment on the system (ES): (a) opening a carbonated beverage; (b) filling a flat tire; (c) a sealed empty gas can expands on a hot day, bowing out the walls.We ordinarily say that U=0 for an isothermal process. Does this assume no phase change takes place? Explain your answer.Check your Understanding Show that QhQh=QcQc for the hypothetical engine of Figure 4.10 The second property to be demonstrated is that all reversible engines operating between the same two reservoirs have the same efficiency. To this, stat with the two engines D and E of Figure 4.10 (a), which are operating between two common heat reservoirs at temperatures Th and Tc . First, assume that D is a reversible engine and that E is a hypothetical irreversible engine that has a higher efficiency than D. If both engines perform the same amount of work W per cycle, it follows from Equation 4.2 that QhQh . It then follows from the first law that QcQc . Figure 4.10 (a) Two uncoupled engines D and E working between the same reservoirs. (b) The engines, With D working reverse. Suppose the cycle of D is so that it operates as a refrigerator, and the two engines are coupled such that the work output of E is used to drive D, as shown in Figure 4.10(b). Since QhQh and QcQc , the net result of each cycle is equivalent to a spontaneous transfer of heat from the cold reservoir to the hot reservoir, a process second law does not allow. The original assumption must therefore be wrong, and it is impossible to construct an irreversible engine such that E is more efficient than the reversible engine D. Now it is quite easy to demonstrate that the efficiencies of all reversible engines operating between the same reservoirs are equal. Suppose that D and E are reversible engines. If they are as shown in Figure 4.10(b), the efficiency of E cannot be greater than the efficiency of D, or second law would violated. If both engines are then reversed, the same reasoning implies that the efficiency of D cannot be greater than the efficiency of E. Combining these results leads to the conclusion that all reversible engines working between same two reservoirs have the same efficiency.
- A great deal of effort, time, and money has been spent in the quest for a so-called perpetual-motion machine, which is defined as a hypothetical machine that operates or produces useful work indefinitely and/or a hypothetical machine that produces mole work or energy than it consumes. Explain, in terms of the first law of thermodynamics, why or why not such a machine is likely to be constructed.The gasoline internal combustion engine operates in a cycle consisting of six parts. Four of these parts involve, among other things, friction, heat exchange through finite temperature differences, and accelerations of the piston; it is irreversible. Nevertheless, it is represented by the ideal reversible Otto cycle, which is illustrated below. The working substance of the cycle is assumed to be air. The six steps of the Otto cycle ale as follows: i. Isobaric intake stroke (OA). A mixture of gasoline and air is drawn into the combustion chamber at atmospheric pressure P0 as the piston expands, increasing the volume of the cylinder from zero to VA . ii. Adiabatic compression stroke (AB). The temperature of the mixture rises as the piston compresses it adiabatically from a volume VA to VB . iii. Ignition at constant volume (BC). The mixture is ignited by a spark. The combustion happens so fast that there is essentially no motion of the piston. During this process, the added heat Q1 causes the pressure to increase from pB to pc at the constant volume VB(=Vc) . iv. Adiabatic expansion (CD). The heated mixture of gasoline and air expands against the piston, increasing the volume from VC to VD . This is called the power stroke, as it is the part of the cycle that delivers most of the power to the crankshaft. v. Constant-volume exhaust (DA). When the exhaust valve opens, some of the combustion products escape. There is almost no movement of the piston during this part of the cycle, so the volume remains constant at VA(=VD) . Most of the available energy is lost here, as represented by the heat exhaust Q2 . vi. Isobaric compression (AO). The exhaust valve remains open, and the compression from VA to zero drives out the remaining combustion products. (a). Using (i)e=W/Q1; (ii)w=Q1Q2; and (iii)Q1=nCv(TCTB),Q2=nCv(TDTA), Show that e=1TDTATCTB. (b). Use the fact that steps (ii) and (iv) are adiabatic to show that e=11r1 where r=VA/VB . The quantity r is called the compression ratio of the engine. (c) In practice, r is kept less than around 7. For larger values, the gasoline-air mixture is compressed to temperatures so high that it explodes before the finely timed spark is delivered. This preignition causes engine knock and loss of power. Show that for r=6 and =1.4 (the value for air), e=0.51 , or an efficiency of 51%. Because of the many irreversible processes, an actual internal combustion engine has an efficiency much less than this ideal value. A typical efficiency for a tuned engine is about 25% to 30%.When a gas expands along AB (see below), it does 500 J of work and absorbs 250 J of heat. When the gas expands along AC, it does 700 J of work and absorbs 300 J of heat. (a) How much heat does the gas exchange along BC? (b) When the gas makes the transmission from C to A along CDA, 800 J of work are done on it from C to D. How much heat does it exchange along CDA?
- An ideal gas expands isothermally along AB and does 700 J of work (see below). (a) How much heat does the gas exchange along AB? (b) The gas then expands adiabatically along BC and does 400 J of work. When the gas returns to A along CA, it exhausts 100 J of heat to its surroundings. How much work is done on the gas along this path?It is unlikely that a process can be isothermal unless it is a very slow process. Explain why. Is the same true for isobaric and isochoric processes? Explain your answer.Does a gas become more orderly when it liquefies? Does its entropy change? If s, does the entropy increase or decrease? Explain your answer.
- A Carnot engine working between two heat baths of temperatures 600 K and 273 K completes each cycle in 5 sec. In each cycle, the engine absorbs 10 kJ of heat. Find the power of the engine.One method of converting heat transfer to doing work is for heat transfer into a gas to take place, which expands, dong work on a piston, as shown in the figure below. (a) Is the heat transfer converted directly to work in an isobaric process, or does it go through another form first? Explain your answer. (b) What about in an isothermal process? (c) What about in an adiabatic process (where heat transfer occurred prior to the adiabatic process)?On an adiabatic process of an ideal gas pressure, volume and temperature change such that pV is constant with =5/3 for monatomic gas such as helium and =7/5 for diatomic gas such as hydrogen at room temperature. Use numerical values to plot two isotherms of 1 mol of helium gas using ideal gas law and two adiabatic processes mediating between them. Use T1=500K,V1=1L, and T2=300K for your plot.