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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?

User Gazreese
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1 Answer

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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
User Aamirl
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