Question

A steel plant operator wishes to change 100 kg of 25℃ iron into molten iron (melting point = 1538℃). How much thermal energy must be added? For iron c=4.5×10^2 J/kg℃ and L_f=2.66×10^5 J/kg.

Answer 1

Given,

The mass of the iron, m=100 kg

The initial temperature of the iron, T₁=25 °C=298.15 K

The melting point of the iron, T₂=1538 °C=1811.15 K

The specific heat of the iron, c=4.5×10² J/kg°C

The latent heat of fusion of iron, L_f=2.66×10⁵ J/kg

The heat needed to raise the temperature of the iron to 1811.15 K is given by,

`\begin{gathered} Q_1=mc\Delta T \\ =mc(T_2-T_1) \end{gathered}`

The heat required to change the phase of the iron from solid to liquid is

`Q_2=mL_f`

Thus the total heat required is,

`\begin{gathered} Q=Q_1+Q_2_{} \\ =mc(T_2-T_1)+mL_f \end{gathered}`

On substituting the known values,

`\begin{gathered} Q=100*4.5*10^2*(1811.15-298.15)+100*2.66*10^5 \\ =94.68*10^6\text{ J} \end{gathered}`

Thus the thermal energy that must be supplied is 94.68×10⁶ J