05.07.2022 - 18:00

# Consider three engines that each use 1650 J of heat from a hot resevoir (temperature = 550 K). These three engines reject heat to a cold resevoir (temperature 330 K). Engine I rejects 1270 J of heat.

Question:

Consider three engines that each use 1650 J of heat from a hot resevoir (temperature = 550 K). These three engines reject heat to a cold resevoir (temperature 330 K). Engine I rejects 1270 J of heat. Engine II rejects 990 J of heat. Engine III rejects 792 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. What is the total entropy change of the universe for (a) engine I, (b) engine II, and (c) engine III? (Note: The total entropy change of the universe is the sum of the entropy changes of the hot and cold resevoirs.)

• The labor market equilibrium can be found by setting the demand for labor equal to the supply of labor in both cities: For Ashton: Demand: Qd = EA = 25,000 - wA/0.0024 Supply: Qs = x, where x is the number of workers willing to work in Ashton at a wage of wA or lower Setting Qd equal to Qs: 25,000 - wA/0.0024 = x Solving for x: x = 25,000 - (20 - wA)/0.0024 x = 8,333.33 + 4166.67wA For Benton: Demand: Qd = EB = 25,000 - wB/0.0004 Supply: Qs = 25,000 - x, where x is the number of workers willing to work in Benton at a wage of wB or higher Setting Qd equal to Qs: 25,000 - wB/0.0004 = 25,000 - x Solving for x: x = wB/0.0004 x = 50,000wB Since the preference for Ashton varies across people, the number of workers in Ashton and Benton will depend on the wage differential between the two cities. Let d represent the wage differential between Ashton and Benton (i.e. d = wB -wA). Using the fact that 40 percent of the working population will choose to work in Benton if the wage differential is $2, we can solve for d: 0.4(25,000) = 10,000 = 2/d d = 0.0002 Now we can solve for the labor market equilibrium: Setting x in Ashton equal to x in Benton and solving for the equilibrium wage differential: 8,333.33 + 4166.67wA = 50,000wB - wB/0.0004 8,333.33 + 4166.67wA = 50,000(wA + d) - (wA + d)/0.0004 Solving for d: d = 0.0001765 Thus, the equilibrium wage differential is$0.0001765 per hour. To find the number of workers employed in both cities and the wage paid in each city, we can use the equations for x in Ashton and Benton: x = 8,333.33 + 4166.67wA = 50,000(wA + d) Plugging in the value for d, we get: x = 50,000wA + 27.94 Thus, the number of workers in Ashton is \approx imately 8,361 and the number of workers in Benton is \approx imately 16,639. Substituting the value of x into the demand equations for each city, we find that the equilibrium wage in Ashton is \approx imately $18.28 per hour, and the equilibrium wage in Benton is \approx imately$18.28 + $0.0001765 =$18.28 + $0.0002 =$18.2802 per hour. Therefore, the labor market equilibrium results in \approx imately 8,361 workers in Ashton earning $18.28 per hour and \approx imately 16,639 workers in Benton earning$18.2802 per hour, with a small wage differential of \$0.0001765 per hour between the two cities.