Question

Problem 5: A monoatomic gas is undergoing a process too quick to live any heat transfer possible to 1/12 of its Original volume. The initial temperature is 20°C, estimate the temperature at the end of the process, Problem 6: A block of stainless steel (m = 3 kgs. To = 50°C) is thrown into a lake whose temperature is 20°C. Cakulate the entropy increase of the universe once the process reaches equilibrium. Problem 7: An engine is producing 200 W from a hot source giving 600 W at 800°C. The engine is releasing heat to the atmosphere at 15°C. What is the effectiveness of this engine? Problem 8: An idealized freezing system is consuming 250 W and transfers 750 W between a cold indoor atmosphere and a hot outdoor atmosphere at 20°C. What is the minimum indoor temperature beyond which the system cannot work anymore for sure?

Answer 1

6o In = 0.33 = 33.3% Given data. 10 2= 1067 mono atomic gas heat transfer is zero too suicky..isentropie procese) undergoing process Volume final I x volume intial- u Initial temperature, TI = 207273 = 293 K. for sentropic process TV 27 = constant 1.697 1-64-1 Tv = Tz Uz 0.67 Tz = final temperature = 293x (10) = 1370.46 K, =1097.46°C. © mass of block, m= 3kg Initial temperature Ti = 50+ 273 = lake temperature T2 = 20 +273 = 2936. (As)universe - Assystem + 323 K @s) surroundings mcln (1/2) + mc (Tr??) Tz = 3x 0.46 in 693) + 3x0.06x(323–293) -0.13452 293 +0.14129 0.00676KTIK Specific heat of Stainless steel is not given so I assumed from data book (p= c = 0.46 kJ/kgk. 1800°C heat engine PPT Goow effectiveness of heat engine workoutput =nof E Ork +200w WE nala o Heat input 150 200

TH 120°C. - COR, = Ideal TL TH-TL Refrigerating effect woric input DME input --> TC 750 W=2506 250 293-TC- 175ow Tu TL = 3(293-TC) TL = 3x293 – 3TL 4TL = 3x 293 C Minimum TL x 293 - 219.15* temperature, Te 219.75 K.=-53-25°C indoor