Chapter 22: inquiries 22:3, 4, 6, 16, 17 troubles 22:1, 2, 6, 7, 8, 9, 11, 28, 31, 32, 41 it is in sure and also do these; do not simply wait and also watch me perform them in class! Q22.3 use the an initial law of thermodynamics to define why the complete energy of one isolated device is always constant.
The very first law that thermodynamics is just a restatement of power Conservation. If a device is secluded no occupational is done on it and no heat is moved to or from it. Power may be transferred from one part of the system to one more but the complete energy continues to be constant.

You are watching: How much heat is extracted from the hot reservoir per cycle?


Q22.4 Is it possible to transform internal energy to mechanical energy? The an initial law of thermodynamics transaction with exactly that. Internal energy can be supplied to transfer power in the because that of warm or in the type of work. Q22.6 In practical heat engines, which execute we have more control of, the temperature that the hot reservoir or the temperature that the cold reservoir? Explain.
The cold temperature is probably going to be ambient atmospheric temperature -- possibly the temperature of a cooling pond or a flow or the ocean. There is little control over any kind of of those. The warm temperature will certainly be the temperature the the burn oil or coal or fissioning Uranium; the is the temperature we can control.
Nearly eliminating friction often means the procedure is almost reversible. A straightforward pendulum is reversible till we look at it closely enough to uncover the friction native the air. Slowly compressing a balloon is virtually reversible.

See more: How Many Bundles Of Shingles Are On A Pallet Of Shingles? Average Weight Of Roof Shingles ( With Examples )


Q22.17 A thermodynamic process occurs in which the entropy of the system alters by - 8.0 J/K. According to the second law the thermodynamics, what deserve to you conclude around the entropy readjust of the environment?
The entropy adjust of the setting -- the surrounding, the remainder of the Universe exterior our "system" -- should be + 8.0 J/K or better so that the full energy the the world does not decrease.
22.1 A warmth engine absorbs 360 J of thermal energy and also performs 25 J of work in each cycle. Find (a) the efficiency of the engine e = 25 J / 360 J e = 0.069 e = 6.9% and (b) the thermal energy expelled in each cycle. 1 - = e = 0.069 Qc/Qh = 1 - 0.069 Qc/Qh = 0.931 Qc/Qh (0.931) Qc = (360 J) (0.931) Qc = 335 J 22.2 A warmth engine performs 200 J of work in each cycle and has an efficiency of 30%. For each cycle, just how much thermal energy is (a) absorbed and Qh = W/e Qh = 200 J/0.30 Qh = 667 J (b) expelled? 1 - = e = 0.30 Qc/Qh = 1 - 0.30 Qc/Qh = 0.70 Qc = Qh (0.70) Qc = (667 J) (0.70) Qc = 467 J 22.6 A specific engine has a power output that 5.0 kW and also an effectiveness of 25%. If the engine expels 8 000 J the thermal power in each cycle, uncover (a) the heat soaked up in each cycle and also 1 - = e = 0.25 Qc/Qh = 1 - 0.25 Qc/Qh = 0.75 Qh = Qc /0.75 Qh = 8 000 J/0.75 Qh = 10 667 J (b) the moment for every cycle.
because that each cycle, W = Qh - Qc W = 10 667 J - 8 000 J W = 2 667 J = 2.667 kJ ns = W / t t = W / p t = 2.667 kJ/5.0 kW t = 0.533 s 22.7 one engine absorbs 1 600 J native a warm reservoir and expels 1 000 J to a cold reservoir in every cycle. (a) What is the efficiency of the engine? e = 1 - < 1 000 J/1 600 J> e = 1 - 0.625 e = 0.375 e = 37.5% (b) What is the power output of the engine if every cycle lasts because that 0.30 s? because that each cycle, W = Qh - Qc W = 1 600 J - 1 000 J W = 600 J then the strength is p = W / t p = 600 J / 0.30 s p = 2 000 W = 2 kW 22.8 A warm engine operates in between two reservoirs in ~ 20oC and 300oC. What is the preferably efficiency feasible for this engine? from our examine of Carnot engines, we know the maximum efficiency is Remember, the course, this temperatures have to be measured in kelvins, Tc = 20oC = 293 K Th = 300oC = 573 K e = 1 - < 293 K/573 K> e = 1 - 0.511 e = 0.489 e = 48.9% 22.9 A power plant operates at a 32% efficiency during the summer once the sea water because that cooling is in ~ 20oC. The plant offers 350oC heavy steam to journey turbines. Presume the plant"s efficiency transforms in the very same proportion as the right efficiency, what is the plant"s efficiency in the winter once the sea water is at 10oC? First, calculation the Carnot efficiencies for summer and also winter, esummer = 1 - <293 K/623 K> esummer = 1 - 0.470 esummer = 0.530 ewinter = 1 - <283 K/623 K> ewinter = 1 - 0.454 ewinter = 0.546 ratio = ewinter /esummer proportion = 0.546/0.530 ratio = 1.03 that is, the Carnot effectiveness in the winter is 1.03 times the Carnot performance in summer. Because of the lower cold temperature, the Carnot performance increases by 3%. If the genuine efficiency transforms by 3% -- indigenous its summer worth of 32% -- climate we intend a winter performance of ewinter /esummer = ratio ewinter /esummer = 1.03 ewinter = 1.03 esummer ewinter = 1.03 (32%) ewinter = 33% 22.11 A power plant has actually been proposed the would manipulate thetemperature gradient in the ocean. The device is to operate between 20oC (surface water temperature) and 5oC (water temperature at a depth of around 1 km). (a) What is the maximum efficiency of such a system? The maximum feasible efficiency is the Carnot efficiency, Tc = 5oC = 278 K Th = 20oC = 293 K e = 1 - < 278 K/293 K> e = 1 - 0.949 e = 0.051 e = 5.1% (b) If the strength output that the plant is 75 MW, how much thermal power is absorbed per hour? First, just how much work-related is done every hour? ns = W/t W = ns t W = (75 MW) (1 hr) W = (75 x 106 W) <(J/s)/W> (1 hr) <3 600 s/hr> W = 2.7 x 1012 J
Now, exactly how much warmth is necessary to provide that lot output work? Qh = W/e Qh = (2.7 x 1012 J) / 0.051 Qh = 5.3 x 1013 J (c) What compensating factor made this proposal that interest despite the worth of the performance calculated in component (a)? there is numerous sea water! The temperature gradient in the ocean provides and also abundant resource of power altho" the small temperature gradient way the effectiveness is an extremely low. 22.14 steam enters a turbine at 800oC and is tired at 120oC. What is the maximum efficiency of this turbine? Tc = 120oC = 393 K Th = 800oC = 1073 K e = 1 - < 393 K/1073 K> e = 1 - 0.366 e = 0.634 e = 63.4% 22.28 What is the adjust in entropy as soon as 1 mole of silver- (108 g) is melted at 961oC? how much warmth is required to melt 108 g the silver? Q = m l Q = (0.108 kg) (8.82 x 104 J/kg) Q = 9.53 x 103 J S = Q/T T = (273 + 961) K T = 1234 K S = (9.53 x 103 J)/(1234 K) S = 7.72 J/K 22.31 calculation the change in entropy that 250 g the water heated slowly from 20oC come 80oC. (Hint: note that dQ = m c dT). This heat is included at different temperatures, for this reason we must take an integral, dQ = m c dT 22.32 An ice cream tray has 500 g that water at 0oC. Calculation the change in entropy of the water as it freezes fully and slowly at 0oC.
Q = m Lf Q = (0.500 kg) (3.33 x 105 J/kg) Q = 1.67 x 105 J S = Q/T S = (1.67 x 105 J)/273 K S = 610 J/K 22.41 an 18-g ice cream cube in ~ 0.0oC is heated until it vaporizes as steam. (a) exactly how much go the entropy increase? just how much heat is required to melt the ice cube into water at 0oC (or 273 K)? Q = m Lf Q = (0.018 kg) (3.33 x 105 J/kg) Q = 599 J That means the entropy added was S = Q/T S = (599 J)/273 K S = 2.2 J/K just how much entropy is included as the temperature increases from 0oC to 100oC (or 373 K)? That"s similar to problem 22.32 that we have currently solved; this, too, of course, calls for an integral. dQ = m c dT just how much warm is forced to vaporize the water into steam at 100oC? Q = m LvQ = (0.018 kg) (2.26 x 106 J/kg) Q = 40,700 J That way the entropy added was

S = Q/T

T = (100 + 273) K

S = (40,700 J)/373 K S = 109 J/K Therefore, the complete entropy adjust is simply the sum of this parts: Stot = <2.2 + 23.5 + 109> J/K Stot = 135 J/K (See Table 20.2) (b) exactly how much energy was required to vaporize the ice cube?
us have currently done (almost) every the work-related (ie, effort) ~ above this, we simply need to pick up the pieces Qtot = Qmelt + Qraise + Qvap Qraise = m c T Qraise = (0.018 kg) (4186 J/kg-K) (100 K) Qraise = 7,535 J Qtot = Qmelt + Qraise + Qvap Qtot = 599 J + 7,535 J + 40,700 J Qtot = 48,834 J
*
*
review Ch23, E-fields go back to ToC (c) Doug Davis, 2002; all legal rights reserved