Question

How much energy does it take to warm up 4.8 kg of iron from room temp (22oC) to 400oC

Answer 1

Given,

Mass of the iron, m=4.8 kg

Initial temperature, T₁=22 °C=295.15 K

Final temperature, T₂=400 °C=673.15 K

The specific heat capacity of the iron is, c=462 J/kg·K

The heat energy required is calculated using the formula,

$Q=mc\Delta T=mc(T_2-T_1)$

On substituting the known values,

$Q=4.8\times462(673.15-295.15)=838.25\text{ kJ}$

Therefore the energy required is 838.25 kJ