Asked by Keru
A 300.0g sample of ice at -30.OC is mixed with 300.0G of water at 90C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capazities of H2O(s) and H20(L) are 2.03 and 4.18 j/g*C respectivly, and the entahply fusion for ice is 6.02 kJ/mol
Answers
Answered by
DrBob222
heat gained by ice from -30 to 0C + heat gained by melting ice + heat gained by water from ice @ 0 C moving to higher T + heat lost by H2O @ 90C = 0
[mass ice x specific heat ice x (Tfinal-Tinitial)] + [mass ice x delta Hfusion] + [mass melted ice x specific heat H2O x (Tfinal-Tinitial)] + [mass warm water x specific heat H2O x (Tfinal-Tinitial)]=0
You have values for everything there except Tf, solve for that.
[mass ice x specific heat ice x (Tfinal-Tinitial)] + [mass ice x delta Hfusion] + [mass melted ice x specific heat H2O x (Tfinal-Tinitial)] + [mass warm water x specific heat H2O x (Tfinal-Tinitial)]=0
You have values for everything there except Tf, solve for that.
Answered by
Anonymous
133.317 cal/grams
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