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Topics include heating neon in a rigid tank under pressure, evaluating water temperature after energy input with isentropic efficiency and heat-loss effects, and estimating minimum electrical power for an adiabatic pump. Additional questions compute entropy change for a reversible heat/work process and total entropy change during quenching of a hot metal part in water, with multiple-choice answers and provided solution steps for selected items.",{"@graph":69,"@context":119},[70,84,106],{"@type":71,"itemListElement":72},"BreadcrumbList",[73,77,79,82],{"item":74,"name":75,"@type":76,"position":8},"https://docshare.wps.com","Home","ListItem",{"item":78,"name":9,"@type":76,"position":14},"https://docshare.wps.com/document/",{"item":80,"name":20,"@type":76,"position":81},"https://docshare.wps.com/document/exam/",3,{"item":83,"name":65,"@type":76,"position":19},"https://docshare.wps.com/document/thermodynamics-practice-problems-questions-9-13/300343/",{"url":83,"name":65,"@type":85,"image":86,"author":91,"headline":65,"publisher":94,"fileFormat":97,"inLanguage":63,"description":67,"dateModified":98,"datePublished":99,"encodingFormat":97,"isAccessibleForFree":100,"interactionStatistic":101},"DigitalDocument",{"url":87,"@type":88,"width":89,"height":90},"https://docshare.wps.com/thumbnails/thermodynamics-practice-problems-questions-9-13/300343.png","ImageObject",300,407,{"name":92,"@type":93},"Aurora","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-11-04","2026-09-18",true,{"@type":102,"interactionType":103,"userInteractionCount":105},"InteractionCounter",{"@type":104},"ViewAction",15,{"@type":107,"mainEntity":108},"FAQPage",[109,115],{"name":110,"@type":111,"acceptedAnswer":112},"How is entropy change calculated for a reversible engine process with given heat loss and work output?","Question",{"text":113,"@type":114},"Compute ΔS using the reversible relation involving heat transfer with the reservoir temperature. The question provides work output and heat loss at a constant temperature to determine the entropy change.","Answer",{"name":116,"@type":111,"acceptedAnswer":117},"How is total entropy change evaluated immediately after quenching a hot metal part in water?",{"text":118,"@type":114},"Sum the entropy changes of the metal and the water using calorimetric assumptions and the entropy formula for sensible temperature change. The provided choices distinguish between entropy increase and decrease, based on heat flow direction.","https://schema.org",{"og:url":83,"og:type":121,"og:title":65,"og:site_name":95,"og:description":67},"article",{"robots":123,"canonical":83},"index,follow",{"doc_id":125,"site_id":62},300343,1790103265,{"code":4,"msg":5,"data":128},{"doc_id":125,"user_id":129,"nickname":92,"user_avatar":130,"doc_module":4,"category_id":19,"category_name":20,"doc_title":65,"doc_description":67,"doc_content":131,"file_id":132,"file_url":133,"file_type":134,"file_size":135,"view_count":105,"is_deleted":4,"is_public":8,"is_downloadable":8,"audit_status":8,"page_count":19,"language":136,"language_code":63,"site_id":62,"html_lang":63,"table_of_contents":137,"faqs":138,"seo_title":139,"seo_description":67,"update_tm":140,"read_time":52,"preview_page_count":4},4810365810221,"https://ap-avatar.wpscdn.com/davatar_155a257f0dc6eb9ab79c44ca47cae57d","|  | Thermodynamics |\n| --- | --- |\n|  |  |\n|  |  |\n\n14-2  F E M E C H A N I C A L P R A C T I C E P R O B L E M S   \n9. 7 kg of neon is stored in a rigid tank at three times atmospheric pressure and a temperature of 70 􀀁 C. 30 kJ of heat is added to the neon. What is most nearly the final temperature of the neon?  \n(A) 70 􀀁 C  \n(B) 73 􀀁 C  \n(C) 74 􀀁 C  \n(D) 77 􀀁 C  \n10. A device expends 130 kJ of energy while pressurizing 10 kg of water initially at 17 􀀁 C. The isentropic efficiency of the device is 50% . Inefficiencies are represented by a heat loss from the device casing. What is most nearly the final temperature of the water?  \n(A) 18 􀀁 C  \n(B) 19 􀀁 C  \n(C) 20 􀀁 C  \n(D) 21 􀀁 C  \n11. An adiabatic pump receives 1.5 kg/s of 15 kPa water and discharges it at 15 MPa. The specific volume of the entering water is 0 .001055 m3/kg. Consider the water to be incompressible. The isentropic efficiency of the pump is 0.82 . Velocity and elevation changes are insignificant. The water does not increase in temperature significantly. Most nearly, what is the minimum electrical power required to drive the pump?  \n(A) 13 kW  \n(B) 20 kW  \n(C) 23 kW  \n(D) 30 kW  \n12. An engine operates at a constant temperature of 90 􀀁 C. Through a reversible process, the engine’s work output is 5.3 kJ, and the heat loss is 4.7 kJ. What is most nearly the change in entropy during the process?  \n(A) 0.013 kJ/K  \n(B) 0.014 kJ/K  \n(C) 0.015 kJ/K  \n(D) 0.016 kJ/K  \n13. In a heat treating process, a 2 kg metal part (specific heat=0.5 kJ/kg􀀃K) initially at 800 􀀁 C is quenched in a tank containing 200 kg of water initially at 20 􀀁 C. What is most nearly the total entropy change of the process immediately after quenching?  \n(A) 2.3 kJ/K (decrease)  \n(B) 0.65 kJ/K (decrease)  \n(C) 0.90 kJ/K (increase)  \n(D) ~~1~~.~~4~~ kJ/K (increase)   \n1.6  \nSOLUTIONS  \n.................................................................................................................................  \n1. The generators are driven by hydraulic turbines. Since the process is adiabatic, and since velocity and elevation changes are insignificant, all of the terms in the steady flow energy equation drop out except work and enthalpy. The turbine work per unit mass is  \nwturbine ¼ hi 􀀄 he  \nIn the absence of compressed water tables (giving enthalpies of subcooled and compressed water), the turbine work must be calculated from more basic principles. Enthalpy is defined as h = p􀀁 + u, so the turbine work per unit mass is  \nwpump ¼ hi 􀀄 he ¼ pi 􀀁i 􀀄 pe 􀀁 e  \nSince the water is incompressible, the specific volume is unchanged. The turbine work per unit mass is  \nwpump ¼ 􀀁i ðpi 􀀄 pe Þ  \nThe power generated by turbines is W_ ¼ m_ 􀀁i ðpi 􀀄 pe Þ  \n¼ 􀀁900 Msg􀀃􀀁0:001 mkg3~~ ~~ 􀀃 ð900 kPa 􀀄 200 kPaÞ  \n¼ 630 MW The answer is (B).  \n2. The first law is  \nQ 􀀄 W ¼ DU  \nQ is the heat transfer into the system across the system boundary. Wis the energy transferred across the system boundary to the surroundings in the form of work. DUis the change in energy stored within the system in the form of internal energy. This expression shows that the energy transferred across the system boundary comes from a change in stored energy.  \nThe answer is (B).  \n3. The energy equation is based on the first law of thermodynamics.  \nThe answer is (C).  \nP P I * [w w w . p p i 2 p a s s . c o m](w w w . p p i 2 p a s s . c o m)   \n L A W S O F T H E R M O D Y N A M I C S  14-5  \n12. The change in entropy is DSout ¼ ~~ ~~TrQ~~ out~~eservoir~~ ~~   \n4:7 kJ ¼  \n90􀀁 C þ 273􀀁¼ 0:013 kJ=K  \nThe answer is (A).  \n13. Calculate the final temperature, Tf.  \nðmcDTÞmetal þðmcDTÞwater ¼ 0 ð2 kgÞ 􀀁0:5 kkgJ􀀃K􀀃 ð800􀀁 C 􀀄 Tf Þ  \nþð200 kgÞ 􀀁4:18 kkgJ􀀃K􀀃 ð20􀀁 C 􀀄 Tf Þ¼ 0  \nTf ¼ 20:93􀀁 C ð21􀀁 CÞ  \nFor a solid or liquid, the total entropy is DS ¼ mc ln ~~ ~~TT~~2~~1  \nSo, considering a system consisting of the metal part and the water in the quenching tank,  \nDSmetal ¼ mc ln ~~ ~~TT~~2~~1  \n¼ ð2 kgÞ 􀀁0:5 kkgJ􀀃K􀀃 ln 21800􀀁􀀁CCþþ273273􀀁􀀁  \n¼ 􀀄1:295 kJ=","cbCaihbPEmejIijT","https://ap.wps.com/l/cbCaihbPEmejIijT","pdf",152018,"English","# Practice Problems\n## Energy and Entropy Questions 9-13\n## Solutions and Calculations","[{\"question\":\"How is entropy change calculated for a reversible engine process with given heat loss and work output?\",\"answer\":\"Compute ΔS using the reversible relation involving heat transfer with the reservoir temperature. The question provides work output and heat loss at a constant temperature to determine the entropy change.\"},{\"question\":\"How is total entropy change evaluated immediately after quenching a hot metal part in water?\",\"answer\":\"Sum the entropy changes of the metal and the water using calorimetric assumptions and the entropy formula for sensible temperature change. The provided choices distinguish between entropy increase and decrease, based on heat flow direction.\"}]","Thermodynamics - Practice Problems - Questions 9-13 | PDF",1789773920]