Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 19, Problem 45P
(a)
To determine
The efficiency of Carnot engine.
(b)
To determine
To prove that no other engine work as a refrigerator between the same two reservoirs can have COP greater than 2.
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Chapter 19 Solutions
Physics for Scientists and Engineers
Ch. 19 - Prob. 1PCh. 19 - Prob. 2PCh. 19 - Prob. 3PCh. 19 - Prob. 4PCh. 19 - Prob. 5PCh. 19 - Prob. 6PCh. 19 - Prob. 7PCh. 19 - Prob. 8PCh. 19 - Prob. 9PCh. 19 - Prob. 10P
Ch. 19 - Prob. 11PCh. 19 - Prob. 12PCh. 19 - Prob. 13PCh. 19 - Prob. 14PCh. 19 - Prob. 15PCh. 19 - Prob. 16PCh. 19 - Prob. 17PCh. 19 - Prob. 18PCh. 19 - Prob. 19PCh. 19 - Prob. 20PCh. 19 - Prob. 21PCh. 19 - Prob. 22PCh. 19 - Prob. 23PCh. 19 - Prob. 24PCh. 19 - Prob. 25PCh. 19 - Prob. 26PCh. 19 - Prob. 27PCh. 19 - Prob. 28PCh. 19 - Prob. 29PCh. 19 - Prob. 30PCh. 19 - Prob. 31PCh. 19 - Prob. 32PCh. 19 - Prob. 33PCh. 19 - Prob. 34PCh. 19 - Prob. 35PCh. 19 - Prob. 36PCh. 19 - Prob. 37PCh. 19 - Prob. 38PCh. 19 - Prob. 39PCh. 19 - Prob. 40PCh. 19 - Prob. 41PCh. 19 - Prob. 42PCh. 19 - Prob. 43PCh. 19 - Prob. 44PCh. 19 - Prob. 45PCh. 19 - Prob. 46PCh. 19 - Prob. 47PCh. 19 - Prob. 48PCh. 19 - Prob. 49PCh. 19 - Prob. 50PCh. 19 - Prob. 51PCh. 19 - Prob. 52PCh. 19 - Prob. 53PCh. 19 - Prob. 54PCh. 19 - Prob. 55PCh. 19 - Prob. 56PCh. 19 - Prob. 57PCh. 19 - Prob. 58PCh. 19 - Prob. 59PCh. 19 - Prob. 60PCh. 19 - Prob. 61PCh. 19 - Prob. 62PCh. 19 - Prob. 63PCh. 19 - Prob. 64PCh. 19 - Prob. 65PCh. 19 - Prob. 66PCh. 19 - Prob. 67PCh. 19 - Prob. 68PCh. 19 - Prob. 69PCh. 19 - Prob. 70PCh. 19 - Prob. 71PCh. 19 - Prob. 72PCh. 19 - Prob. 73PCh. 19 - Prob. 74PCh. 19 - Prob. 75PCh. 19 - Prob. 76PCh. 19 - Prob. 77PCh. 19 - Prob. 78PCh. 19 - Prob. 79PCh. 19 - Prob. 80PCh. 19 - Prob. 81PCh. 19 - Prob. 82PCh. 19 - Prob. 83PCh. 19 - Prob. 84PCh. 19 - Prob. 85PCh. 19 - Prob. 86PCh. 19 - Prob. 87PCh. 19 - Prob. 88PCh. 19 - Prob. 89P
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- Check Your Understanding A Carnot engine operates between reservoirs at 400 and 30 . (a) What is the efficiency of the engine? (b) If the engine does 5.0 J of work per cycle, how much heat per cycle does it absorb from the high-temperature reservoir? (c) How much heat per cycle does it exhaust to the cold-temperature reservoir? (d) What temperatures at the cold reservoir would give the minimum and maximum efficiency?arrow_forwardDoes the entropy increase for a Carnot engine for each cycle?arrow_forwardA Carnot engine operates between 550 and 20 baths and produces 300 kJ of energy in each cycle. Find the change in entropy of the (a) hot bath and (b) cold bath, in each Carnot cycle?arrow_forward
- A Carnot engine is used to measure the temperature of a heat reservoir. The engine operates between the heat reservoir and a reservoir consisting of water at its triple point. (a) If 400 J per cycle are removed from the heat reservoir while 200 J per cycle are deposited in the triple-point reservoir, what is the temperature of the heat reservoir? (b) If 400 J per cycle are removed from the triple-point reservoir while 200 J per cycle are deposited in the heat reservoir, what is the temperature of the heat reservoir?arrow_forwardA Carnot engine has an efficiency of 0.60. When the temperature of its cold reservoir the efficiency drops to 0.55. If initially Tc=27, determine (a) the constant value of Th and (b) the final value of Tc.arrow_forwardCheck Your Understanding A 50-g copper piece at a temperature of 20 is placed into a large insulated vat of water in 100 . (a) What is the entropy change of the copper piece when it reaches thermal equilibrium with the water? (b) What is the entropy change of the water? (c) What is the entropy change of the universe?arrow_forward
- To increase the efficiency of a Carnot engine, should the temperature of the hot reservoir be raised or lowered? What about the cold reservoir?arrow_forwardAn engine is found to have an efficiency of 0.40. If it does 200 J of work per cycle, what are the corresponding quantities of heat absorbed and discharged?arrow_forwardCalculate the net work output of a heat engine following path ABCDA as shown below.arrow_forward
- Check Your Understanding A quantity of heat Q is absorbed from a reservoir at a temperature Th by a cooler reservoir at a temperature Tc . What is the entropy change of the hot reservoir, the cold reservoir, and the universe?arrow_forwardA 0.50-kg piece of aluminum at 250 is dropped into 1.0 kg of water at 20 . After equilibrium is reached, what is the net entropy change of the system?arrow_forwardThe 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%.arrow_forward
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