2 kg of iron at 80°C is placed in a copper can, mass 0.5 kg, containing 1 kg of water at 20°C. After stirring, the temperature of the mixture is 30°C. Calculate the specific heat capacity of iron.

438 J/kg/°C

Explanation:

Heat lost by iron = heat absorbed by water/copper

-m₁C₁ΔT₁ = m₂C₂ΔT₂ + m₃C₃ΔT₃

- (2 kg) C (30°C - 80°C) = (1 kg) (4186 J/kg/°C) (30°C - 20°C) + (0.5 kg) (385 J/kg/°C) (30°C - 20°C)

(100 kg°C) C = 41860 J + 1925 J

C = 438 J/kg/°C