Latent Heat of Fusion Formula - GeeksforGeeks

As a result, the latent heat of fusion includes the process of adding heat to melt a solid. Its formula is given by calculating the total amount ... 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ImproveArticle SaveArticle LikeArticle LastUpdated: 11May,2022ReadDiscussViewDiscussion ImproveArticle SaveArticle LikeArticle Latentheatoffusionisdefinedastheamountofheatreceivedbyasolidbodytotransformitintoaliquidwithoutadditionaltemperature.Itcanbeinterpretedasthephasetransitionbetweensolidandliquidphases.Heatreferstothepassageofheatenergybetweentheobjectsinthiscontext.Asaresult,thelatentheatoffusionincludestheprocessofaddingheattomeltasolid.Itsformulaisgivenbycalculatingthetotalamountofheatabsorbedbythebody.ItisdenotedbythesymbolQ.Itsunitofmeasurementiscalories(Cal)anddimensionalformulaisgivenby[M0L2T−2].Formula Q=mL+mcΔtwhere,Qisthelatentheatoffusion,misthemassofbody,Listhespecificlatentheatoffusion,cisthespecificheatofbody,Δtisthetemperaturechangeduringheatabsorption.SampleProblemsProblem1.Calculatethelatentheatoffusionforabodyofmassof40gat20°Cifitabsorbsheatat80°C.Itisgiventhatthespecificlatentheatofsteamis540cal/gandthespecificheatofthebodyis0.5cal/g°C.Solution:Wehave,m=40Δt=80–20=60L=540c=0.5Usingtheformulaweget,Q=mL+mcΔt=(40×540)+(40×0.5×60)=21600+1200=22800calProblem2.Calculatethelatentheatoffusionforabodyofmass20gat40°Cifitabsorbsheatat100°C.Itisgiventhatthespecificlatentheatofsteamis540cal/gandthespecificheatofthebodyis0.1cal/g°C.Solution:Wehave,m=20Δt=100–40=60L=540c=0.1Usingtheformulaweget,Q=mL+mcΔt=(20×540)+(20×0.1×60)=10800+120=10920calProblem3.Calculatethelatentheatoffusionfor7gofwatergettingconvertedintoice.Itisgiventhatthespecificlatentheatoficeis80cal/g.Solution:Wehave,m=7L=80Usingtheformulaweget,Q=mL=7(80)=560calProblem4.Calculatethelatentheatoffusionfor60gofsteamgettingconvertedintowater.Itisgiventhatthespecificlatentheatofwateris533cal/g.Solution:Wehave,m=60L=533Usingtheformulaweget,Q=mL=60(533)=31980calProblem5.Calculatethemassofwatergettingconvertedintoicegiventhatthespecificlatentheatoficeis80cal/gandthelatentheatoffusionis200cal.Solution:Wehave,Q=200L=80Usingtheformulaweget,Q=mL=>m=Q/L=>m=200/80=>m=2.5gProblem6.Calculatethemassofsteamgettingconvertedintowatergiventhatthespecificlatentheatofwateris533cal/gandthelatentheatoffusionis700cal.Solution:Wehave,Q=700L=533Usingtheformulaweget,Q=mL=>m=Q/L=>m=700/533=>m=1.31gProblem7.Calculatethelatentheatoffusionforabodyofmass30gifitsspecificlatentheatofsteamis540cal/gandheatabsorbedbyitis200calories.Solution:Wehave,m=30L=540Q’=200Usingtheformulaweget,Q=mL+Q’=(30×540)+200=16200+200=16400calMyPersonalNotes arrow_drop_upSave LikePreviousSustainabilityofDevelopmentNext LithiumChlorideFormula-Structure,Properties,Uses,SampleQuestionsRecommendedArticlesPage:12,May2227,May2116,Apr2201,Mar2225,Jan2227,Jan2208,Feb2211,Apr2230,Jan2201,Mar2207,Mar2207,Mar2211,Apr2224,Apr2225,Apr2225,Apr2212,Mar2222,Jun2115,Jun2120,Jun2126,Jun2101,Aug2106,Mar2227,May21ArticleContributedBy:jatinxcx@jatinxcxVotefordifficultyEasy Normal Medium Hard ExpertArticleTags:Physics-FormulasPickedPhysics-MAQSchoolLearningSchoolPhysicsReportIssueWritingcodeincomment? 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