4.60E6 J = [mass ice x specific heat ice x (Tfinal-Tinitial)] + [mass ice x heat fusion] + [mass water from ice x specific heat water x (Tfinal-Tinitial)]
Solve for mass ice/water
Watch that first (Tfinal-Tinitial) because it can be tricky. Tfinal = 0 and Tinitital = -6.1; therefore, it is 0 -(-6.1) = 0+6.1
How many grams of ice at -6.1C can be completely converted to liquid at 27.4C if the available heat for this process is 4.60×10^3 kJ ?
For ice, use a specific heat of 2.01J/(g*C) and traingle Hfus = 6.01kJ/mol.
2 answers
figure it out
2 answers