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