How many joules are needed to warm 875-grams of iron, specific heat 0.448 j/g°c, from 25.0°c to 345°c?

When heat (q) is absorbed by "m" grams of a substance then the change in temperature is given as,

q = m Cp (T₂ - T₁) - - - (1)

where;

Cp = Specific Heat

Specific heat of Iron is 0.448 J/g.°C

Initial Temperature = T₁ = 25 °C

Final Temperature = T₂ = 345 °C

Mass of Iron = m = 875 g

Putting values in eq. 1,

q = 875 g * 0.448 J/g.°C * (345 °C - 25 °C)

q = 1254400 J

Or,

q = 1.25 * 10⁶ J