Question

Using information in Appendices B and C in the textbook, calculate the minimum number of grams of propane, C3H8(g), that must be combusted to provide the energy necessary to convert 5.75 kg of ice at -25.0 ∘C to liquid water at 79.0 ∘C.

Answer 1

Enthalpy of combustion of propane, ΔHoC = –2202.0 kJ/mol

heat required(q) = mass of ice(m)*Sice*DT1 + m*DHfus + mass of water(m)*Swater*DT2

                 = 5.75*10^3*2.087*(0 - (-25))+(5.75*10^3/18)*6.01*10^3 + 5.75*10^3*4.184*(79-0)
            
                = 4120.45 kj

no of mol of propane required (n) = q/DH0c = 4120/2202 = 1.871 mol

   mass of propane, C3H8(g) required = n*Mwt = 1.871*44 = 82.324 grams