Latent Heat of Fusion Formula with Solved Examples

The heat absorbed by the substance, or the latent heat of the fusion formula, is represented as when m kg of solid changes to a fluid at a fixed temperature ... Getfreeliveclasses&testontheapp Download + Home»Formulas»PhysicsFormulas»LatentHeatOfFusionFormula LatentHeatofFusionFormula latentheatoffusionformula:Exploremoreaboutthelatentheatoffusionformulawithsolvedexamples. TableofContent   latentheatoffusionformulaLatentheatoffusionistheamountofheatreceivedbyasoliditemtotransformitintoaliquidwithoutincreasingthetemperaturefurther.BecauseseaiceandbrinemaylivetogetheratanytemperatureandmeltatatemperatureotherthanoCwhenbathedinaconcentratedsaltsolution,theamountoflatentheatiscomplicatedinthecaseofseaice,muchasitisinthewallsofbrinecellswhenbrinecellsmigrate.Theheatabsorbedbythesubstance,orthelatentheatofthefusionformula,isrepresentedaswhenmkgofsolidchangestoafluidatafixedtemperature,whichisitsmeltingpoint.L=m* Ldenotesthesubstance’sparticularlatentheatoffusion.TheheatthatthematerialabsorbsorreleasesisrepresentedasQ=mctasthetemperatureofthesubstancevariesfromt1(lowtemperature)tot2(hightemperature).mc(t2–t1)=QQ=mL+mcΔtisthetotalquantityofheatabsorbedoremittedbythesubstance.Example1.At20°C,apieceofmetalweighs60g.0.5gofthesteamcondensesonitwhenitissubmergedinasteamstreamat100°C.Giventhatthelatentheatofsteamis540cal/g,calculatethespecificheatofthemetal.Solution:Letcbethemetal’sspecificheat.Themetalhasgatheredheat.Q=mcΔtQ=60×c×(100–20)Q=60×c×80calThesteam’sheatisexpelled.Q=m×LQ=0.5×540calAccordingtotheconceptofmixing,Theamountofheatprovidedisequaltotheamountofheatabsorbed.0.5×540=60×c×80c=0.056cal/g°C2.If64500caloriesofheatareextractedfrom100gofsteamat100degreesC,calculatetheamountofwatertransformedtoice.Thelatentheatoficeandsteam,respectively,is80calpergramand540calpergram.Solution:Thetemperatureofthewaterwilldropfrom100degreesCto10degreesCifthecompletesteamistransformedintowaterat100degreesC.Also,aportionofthewaterat0degreesCforiceformation.Theheatrequiredtoconvertsteamat100degreesCelsiustowaterat100degreesCelsius=mLsteam=100540cal=54000calories.Thewatercontentis6.25grams. Frequentlyaskedquestions GetanswerstothemostcommonqueriesrelatedtotheLatentHeatofFusionFormula. Whatistheformulaforcalculatinglatentheatoffusion? Ans.Theheatabsorbedbythesubstance,orthelatentheatofthefusionformula,isrepresentedaswhenmkgofsolidchangestoafluidatafix...Readfull Giveanexampleoflatentheatoffusion. Ans.Thelatentheatoffusionofonekilogramofwater,forexample,is333.55kilojoules,whichistheamountofh...Readfull Ans.Theheatabsorbedbythesubstance,orthelatentheatofthefusionformula,isrepresentedaswhenmkgofsolidchangestoafluidatafixedtemperature,whichisitsmeltingpoint.Q=m* L Ans.Thelatentheatoffusionofonekilogramofwater,forexample,is333.55kilojoules,whichistheamountofheatenergythatmustbegiventotransform1kgoficewithoutaffectingtheenvironment’stemperature(whichisheldatzerodegreesCelsius). ImportantMathsFormulas Trigonometryformula Percentageformula Simpleinterestformula Distanceformula Standarddeviationformula Meanformula Areaofasquareformula Modeformula ImportantPhysicsFormulas Accelerationformula Powerformula Velocityformula Averagespeedformula Pressureformula Momentumformula Workformula Displacementformula ImportantChemistryFormulas Chemicalformula Ureaformula Bleachingpowderformula Molarityformula Ammoniaformula Ethanolformula Oxalicacidformula Acetoneformula DownloadJEEMain2022QuestionPapers . Explore AFCAT APEAMCET BankExam BPSC CAFoundation CAPF CAT CBSEClass11 CBSEClass12 CDS CLAT CSIRUGC GATE IITJAM JEE KarnatakaCET KarnatakaPSC KeralaPSC MHTCET MPPSC NDA NEETPG NEETUG NTAUGC RailwayExam SSC TSEAMCET UPSC WBPSC Sharevia COPY