CH301 - icewatersteam - The Ice, Water, and Steam Problem...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
HEAT added temperature ice steam water 100 g 100 g 12 g Pull all components to 0˚C. The water is already at 0˚C. The ice just needs to be melted. The steam must be cooled, condensed, and then cooled again. 2.03 J/g˚C · 12 g · 10˚C = 243.6 J 2260 J/g · 12 g = 27120.0 J 4.184 J/g˚C · 12 g · 100˚C = 5020.8 J 32384.4 J 334 J/g · 100 g = 33400.0 J 33400.0 J difference = 1015.6 J q = H · m m = q/ H m = 1015.6 / 334 m = 3.0 grams of ice solve for m COLD SIDE HOT SIDE cold side wins! This heat is equivalent to the mass of ice that did NOT get melted by the
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online