Melting requires absorption of heat to bring a substance from the solid
to the liquid state. But before that can happen, the temperature of the
substance must be raised first from its initial temperature to its melting
point temperature, and that requires another amount of heat.
The melting point of aluminum is at Tm = 660.32°C.
So first, the temperature must be raised from the initial value To = 26.4°C
to the final value at Tm. That means a change in temperature of
ΔT = Tm – To = 633.92°C
With M = 1000*(0.014 kg) = 14 kg = 1.4(e+04) g and C = 0.897 J/(g*K),
the amount of heat needed to do that is given by
(1) ..... ΔQ = M*C* ΔT = [1.4(e+04) g]*[ 0.897 J/(g*K)]*( 633.92°C) = 7.96(e+06) J.
Note that the formula in (1) applies only whenever there is no change in
the state of material taking place.
Let us call the heat absorbed in (1) by ΔQ1. Then ΔQ1 = 7.96 MJ
where 1 MJ = 1.0(e+06) J.
Now, to bring the solid at its melting point temperature to its liquid state still
at that melting point temperature, the other amount of heat required is
(2) ..... ΔQ2 = M*Lf = [1.4(e+04) g]*[ 397 J/g ] = 5.56(e+06) J = 5.56 MJ
where Lf = 397 J/g is the heat of fusion of aluminum (the heat needed to bring
1.0 g of aluminum from the solid state at its melting point temperature to the
liquid state, still at that temperature).
It follows that the total amount of heat needed to melt 14 kg of aluminum is
ΔQ = ΔQ1 + ΔQ2 = 13.52 MJ