calculate the minimum number of grams of propane, C3H8(g), that must be combusted to provide the energy necessary to convert 6.05 kg of ice at -24.0 ∘C to liquid water at 71.5 ∘C.

Answer 1

Look up the heat of combustion of propane (j/g). I think it is 50kj/gram
mass(50kj/g)=6.05kg(cice)(24)+6.05(Hfice)+6.05(cwater)(71.5)
solve for mass.