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Solution: An ideal vapor-compression refrigeration cycle

Thermodynamics: An Engineering Approach | 8th Edition | ISBN: 9780073398174 | Authors: Yunus A. Cengel ISBN: 9780073398174 171

Solution for problem 1113 Chapter 11

Thermodynamics: An Engineering Approach | 8th Edition

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Thermodynamics: An Engineering Approach | 8th Edition | ISBN: 9780073398174 | Authors: Yunus A. Cengel

Thermodynamics: An Engineering Approach | 8th Edition

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Problem 1113

An ideal vapor-compression refrigeration cycle that uses refrigerant-134a as its working fluid maintains a condenser at 800 kPa and the evaporator at 2128C. Determine this systems COP and the amount of power required to service a 150 kW cooling load.

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Biological Radiation ­ Increase in radiation, break chemical bonds­> ionizing radiation ­ Can cause malfunction of cells Acute Effects ­ Increase in radiation caused weakened immune system, decrease in nutrient absorption Chronic ­ DNA mutated, genetic defects, sterilization Measuring Radiation ­ Curie (Ci) 3.7x10^10 ­ Gray (Gy) amount of Energy absorbed by body tissue from radiation 1 Gy= 15/kg body tissue ­ Rad 1 rad=0.01 kg ­ A correction factor account for effect the result of RBE, rems, rads (RBE)=rems roentgen equivalent man ­ 100 rem acute radiation syndrome Factors determine Biological radiation 1. Increase in energy, larger effect 2. Better the ionizing effect gamma>>beta>alpha 3. Ionizing larger effect alpha>beta>gamma\ 4. Radioactive half life 5. Biological half life of element 6. Physical state of material Biological Affects ­ 20­100 rems decease white blood cell count , increase cancer risk 100­400 radiation sickness, increase in cancer risk 500+ death Radioisotopes­Medicine ­ Radiotracers tagged isotopes, cause half life use decrease ionizing (beta and gamma) measure body function ­ PET scan: F­10 emitter size and stage of tumors, metabolisms and function,brain scan ­ Radiotherapy: cancer treatment Brachytherapy, teletherapy, radiopharmaceutical therapy Bone Scans ­ Find bone cancer, metastasized cancer to bone, abnormal bone structure Nonmedical radioactive isotopes ­ Smoke detectors ­ Am­241 smoke blocks ionized air, breaks circuit ­ insect control­ sterilize males ­food preservative ­radioactive tracers followed tagged atom in reaction ­chemical analysis (NAA) ­authenticate art ­crime scene investigation ­measure properties of materials: corrosion, track flow, increase temp, defects in pipelines 1 ­Nonstick pans ­sterilize makeup, hair products and contact lenses Ch 23 Inorganic Chemistry Gemstones ­due to the presence of Cr3+­the difference lies in the crystal hosting the ion, ruby Al3+ Al2O3 Cr3+ Al #= ions BeAl2(SiO3) emerald Properties and e­ configuration of transition metals ­ Differ than main group metals ­ Increase melting increase density ­ 2 valence e­ E­ configurations ­1st and 2nd ns^2(n­1)d^x ­1st=[Ar]4s^23d^x; 2nd[Kr]5s^24d^x ­3rd and 4th ns^2(n­2)f^13(n­1)dx ­deviate from pattern s­> d group 1B complete subshell ­ions loose ns e­ first (n­1)d ­e­ leaves from s first Example 1: ​ Zr=[Kr]5s^24d^2 Co^3+=[Ar]4s^23d^7 Ni=[Ar]4s^23d^8 paramagnetic Te^2+=[Kr]5s^24d^5 W=[Xe]6s^24f^145d^4 Irregular electron configurations ­ 4s has less energy than 3d 4s fills before 3d ­ Difference is small ­ Must be found experimentally ­ Cr 4s^13d^5 Cu 4s^23d^10 Mo 5s^14d^5 Ru5s^14d^7 Pd 5s^04d^10 Atomic size ­radii of transition metals are similar, small increase in size down a column ­3rd series=2nd size ­lanthanide series decrease in size, 3rd transitions atoms after lanthanides 2nd&3rd same ­ F orbital is not penetrating, do not shield valence e­ Zeff>e­ shielding ­ Pull from nucleus, greater Zeff stronger pull same size Ionization energy ­ First increase across a series ­ 1st ionization 3rd series> than 1st and 2nd ionization E, valence e­ held tightly Zeff ­ Opposite main group elements Electronegativity ­increase across a series, except last element which has a filled d orbital and cannot accept anymore e­ 2 ­Increase between 1 and 2, 2=3 opposite main group elements Oxidation states ­ Have 2 or more, increase in oxidation state from 3B to 7B Complex Ions ­ Monoatomic cations combine with monoatomic anions or neutral molecules ­ Attached anions or neutral molecules are ligands (lone pair e­) ­ + on complex ion can be + or 0 depends on number of ligands attached Coordination compounds (Lewis acid, base reactions) ­complex ions with counter ions­ neutral compound ­ primary valence oxidation number of metal ­secondary valence number ligands bonded to metal (coordination number) ­coordination number can range from 2­12 common is 6 and 4 Complex Ion formation ­base= ligand ­bond formed is a coordinate covalent bond Ligand with extra teeth ­density ­ligands form more than one bond ­lone pairs separate both ligands to metal ­chelate complex in multidentate ligand (chelating agent) Naming Coordination Compounds 1. Name non complex ion 2. Determine ligand names and list in alphabetical order 3. Determine name of metal cation 4. Name complex in b: alphabetically number found in complex ion 5. Name cation followed by anion 3 exam questions on next exam Example 3: ​[Cr(H2O)3Cl]2+ ​ pentaaquochlorochromium(III)chloride K3[Fe(CN)6]^3­ ​ potassium hexcyanoferrate(III) K2CuI4​ potassium tetraiodocuperate(II) [Cu(NH3)2(CO)4](NO3)3 ​ diamminotetracarbonylcobalt(III)nitrate Isomers ­ Structural isomers same number and types of atoms attached in a different order ­ Stereoisomers same order atoms or groups point in different directions Linkage isomers ­ Structure isomers that have ligands attached to the central atom through different ends of the ligand structure Geometric isomers ­stereoisomers ­cis trans two identical ligands adjacent (cis) opposite (trans) in the structure ­cis trans isomers in the square planar complexes MA2B2 ­cis trans octahedral MA4B2 3 ­fac mer isomer has 3 identical ligands in an octahedral complex either are adjacent to each other making 1 face (fac) or form an arc around the center (mer) structure­ octahedral MA3B Example 4: ​see powerpoint for structures and explanations Optical isomers ­ Stereoisomers, nonsuperimposable mirror images of each other ­ Left (s) sinister, right R(restere) ­ amino acids levaratory (L) right dectoraratory (R ) Example 5: ​see powerpoint for images, nonsuperimposable, cis is optically active, trans is not optically active Example 6: ​see powerpoint for images, rotate 180 degrees symmetrical not optically active Bonding in Coordination compounds: Valence bond theory ­bond filled orbital or ligand overlaps an empty atomic orbital on metal ion ­does not explain color or magnetic properties Crystal Field Theory ­ Bonds form due to the attraction of the e­ on the ligand for the charge on metal cation ­ Depending on direction of the ligand, charge orientation of unhybridized orbitals, there are different electron electron repulsions ­ This leads to a splitting of the energies of the d orbitals ­ Difference in energy depends on complex formed and bonds of ligands ­ Crystal field splitting electron, strong field split and weak field split of the ligands Crystal field splitting ­ Ligands in octahedral complex same space as d lobes ­ Repulsions between electron pairs in bonds and any potential electrons and orbitals increasing in energy of orbitals complex and metal ion and oxidation state ­ X2 x2­y2 strong xy yz xz degenerate Color and complex ions ­ Transitions metals cause color in crystals and solutions ­ Color depends on energy change ­ Observed color is the complementary color of the absorbed color of the complex ion due to electronic transitions between electrons Ephoton=hv=hc/lambda=deltaoct On exam ion will be in kJ/mol remember to multiply by avagardo’s number Example 7: b​ lue [Cu(NH3)6]2+ ­> orange 615 nm Delta oct=E=hc/lambda=(6.626x10^­43)(3x10^8)/615x10^­9 =3.23x10^­19 J (6.022x10^23)= 195 kJ/mol Example 8: ​red [Fe(CN)6]3→ green 525 nm (6.626x10^­34)(3x10^8)/525x10^­9=3.79x10^­19 J/ion (6.022x10^23)=2 ​ 28 kJ/mol Ligands and crystal field strength ­ Size of energy group depends on what kind of ligands are attached to transition metal ion ­ Strong split CN­>NO2­>en>NH3 ­ Weak H2O>OH­>F­>Cl­>Br­ ­ Recognize strong vs weak 4 ­size of energy gap depends on type of cation, increases as charge on metal cation increases Co3+>Cr3+>Fe3_>Fe2+>Co2+>Ni2+>Mn2+ Magnetic properties and crystal field strength ­electron configuration of metal ion with split d orbitals depends on strength 4th and 5th increase energy if split weak energy small and unpaired and paramagnetic X2 x2­y2 high spin ­4­6th pair dxx dyx dxz strong energy gap large paired e­ diamagnetic low spin ­Increase decrease spin only affects d4­d7 Questions similar to those below will be on exam Example 9: ​see powerpoint Example 10: ​ see powerpoint Chrome complexes ­color depends on strength of attached ligands ­stronger the crystal field the shorter the lambda Tetrahedral geometry and crcystal spin ­ligands interact more stronger with planar orbitals in tetrahedral, increase in energy ­Reverse order compared to octahedral ­see powerpoint for diagram ­increase in spin increase in split deltatet=4/9deltaoct On exam do not have to memorize crystal fields tell the difference of one orbital setting vs the other, what is high what is low, etc Linear See powerpoint Colors of rubies and emeralds ­ Rubies Cr3+ 6 O2­ octahedral Al2O3 crystal lattice ­ Emerland Be3Al2 (SiO3) 6O2­ ­ Cr2+ e­ [A]3d3 ­ High split, high energy ruby spectroscopy Applications ­extraction of metal from ores: Au and Ag cyanide complexes, nickel ­chelate lead: EDTA for Pb poisoning ­chemical analysis: ble CosCN+ red FeSCN+, nickel and Pd2+ from insoluble colored precipitates ­paint prussian blue ­biomolecules: porphyrin, cytochrome C, hemoglobin, chlorophyll ­Cubric anhydrase: acidity of blood ­drugs and therapeutic agents: anticancer drugs 5

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Chapter 11, Problem 1113 is Solved
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Textbook: Thermodynamics: An Engineering Approach
Edition: 8
Author: Yunus A. Cengel
ISBN: 9780073398174

The full step-by-step solution to problem: 1113 from chapter: 11 was answered by , our top Engineering and Tech solution expert on 12/23/17, 04:44PM. Thermodynamics: An Engineering Approach was written by and is associated to the ISBN: 9780073398174. The answer to “An ideal vapor-compression refrigeration cycle that uses refrigerant-134a as its working fluid maintains a condenser at 800 kPa and the evaporator at 2128C. Determine this systems COP and the amount of power required to service a 150 kW cooling load.” is broken down into a number of easy to follow steps, and 40 words. This textbook survival guide was created for the textbook: Thermodynamics: An Engineering Approach, edition: 8. This full solution covers the following key subjects: . This expansive textbook survival guide covers 17 chapters, and 2657 solutions. Since the solution to 1113 from 11 chapter was answered, more than 554 students have viewed the full step-by-step answer.

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Solution: An ideal vapor-compression refrigeration cycle