HEAT addedtemperatureicesteamwater100 g100 g12 gPull all components to 0˚C. The water is already at 0˚C. The ice just needs to be melted. Thesteam must be cooled, condensed, and then cooled again.2.03 J/g˚C · 12 g · 10˚C=243.6 J2260 J/g · 12 g= 27120.0 J4.184 J/g˚C · 12 g · 100˚C=5020.8 J32384.4 J334 J/g · 100 g= 33400.0 J33400.0 Jdifference = 1015.6 Jq = ∆H · mm = q/∆Hm = 1015.6 / 334m = 3.0 grams of icesolve for mCOLD SIDEHOT SIDEcold side wins! This heat is equivalent to themass of ice that did NOT get melted by the
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This note was uploaded on 11/02/2008 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.