q needed to raise T of ice at -24.3 to zero is q1 = mass ice x specific heat ice x (Tfinal-Tinitial). We don't know mass ice yet but this equation would look like this.
q1 = mass ice x 2.01 x [0-(-24.3)] = mass ice x 48.84 = 48.84m
q2 = heat needed to melt ice at zero to liquid water at zero.
q2 = mass ice x heat fusion. The heat fusion is given in the probem as 6.01 kJ/mol and that needs to be converted to J/g. That number is 334 J/g*C. We don't know mass ice yet. The equation would look like this.
q2 = mass ice x 334 J/g = 334m.
q3 = head need to raise T of liquid water at zero C to 23.9 is
q3 = mass H2O x specific heat H2O x (Tfinal-Tinitial). We don't know mass liquid water but it will be the same as mass ice. Substitute as I've done for q1 and q2 to find q3 in terms of m.
Add q1 + q2 + q3 = total q needed in terms of mass ice and set that = 4820000 (the available heat) and solve for m = mass ice in grams.
Post your work if you get stuck.
How many grams of ice at -24.3∘C can be completely converted to liquid at 23.9∘C if the available heat for this process is 4.82×103kJ ?
For ice, use a specific heat of 2.01J/(g⋅∘C) and ΔHfus=6.01kJ/mol.
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