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
This is the end of the preview. Sign up
access the rest of the document.