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i Safari File Edit View History Bookmarks Window Help Homework 14 | | + |www.webassign.net/web/Student/Assignment-Responses/submit?dep= 11 594598 Reader Elegant and...eddinglooks Wedding Hai...irstyles HQ Wedding Hai...dding Ideas half up half Syera Sites White Lace ...nedress.com Model Safety...cation Blog Homework 14 Chegg Study | Guided Solutions and Study Help Chegg.com PM 10. + -16 points CJ8 15.P.075 My Notes Consider three engines that each use 2240 J of heat from a hot reservoir (temperature = 560 K). These three engines reject heat to a cold reservoir (temperature = 340 K). Engine I rejects 1360 J of heat. Engine II rejects 655 J of heat. Engine III rejects 1479 J of heat. One of the engines operates reversibly, and two operate irreversibly. However, of the two irreversible engines, one violates the second law of thermodynamics and could not exist. For each of the engines determine the total entropy change of the universe, which is the sum of the entropy changes of the hot and cold reservoirs. On the basis of your calculations, identify which engine operates reversibly, which operates irreversibly and could exist, and which operates irreversibly and could not exist. (Round off to two decimal places.) AS (J/K) Engine I Engine II Engine III Type of operation Select Select Select Submit Answer Save Progress Practice Another Version 11. + -13 points CJ815.P.077 My Notes (a) Find the change in entropy of the H20 molecules when 3.66 kilograms of ice melts into water at 273 K. J/K (b) Find the change in entropy of the H20 molecules when 3.66 kilograms of water changes into steam at 373 K. J/K (c) On the basis of the answers to parts (a) and (b), discuss which change creates more disorder in the collection of H20 molecules

Explanation / Answer

  In all cases, the hot reservoir transfers 2240J of thermal energy to the engine at a temperature of 560K. The change in thermal energy of the hot reservoir, q_hot, is -2240J, and the change in entropy of the hot reservoir is S = q/T = (-2240J)/(560K) = -4 J/K.

The engines add thermal energy to the cool reservoir at at temperature of 340K. The entropy changes of the cool reservoir due to the three engines is therefore:

Engine 1: q/T = (1360J)/(340K) = +4 J/K
Engine 2: q/T = (655J)/(340K) = +1.93 J/K
Engine 3: q/T = (1479J)/(340K) = +4.35 J/K

The change in entropy of the universe is the sum of the changes in the hot and cold reservoirs:


Engine 1: S = (4 - 4.00)J/K = 0 J/K
Engine 2: S = (1.93 - 4.00)J/K = -2.07 J/K
Engine 3: S = (4.35 - 4.00)J/K = 0.35 J/K

The first engine operates reversibly (the change in entropy of the universe is zero). The second engine violates the 2nd law because the entropy of the universe would decrease. It is therefore not a physically realizable engine.

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