Question

A piece of iron of unknown mass has an initial temperature 210∘C. It is dropped into an aluminum container of mass 0.2 kg containing 1 litre (1.0 kg) of water both of which are at a temperature of temperature 20 ∘C . The final equilibrium temperature of the system when energy transfer between the iron and the water finally stops is 28.5 ∘C. (Assume no thermal energy gets lost.) What is the mass of the iron piece? Express your answer in kg to two decimal places.

The specific heat of iron is 450 J/kg ∘C, the specific heat of aluminum is 920 J/kg ∘C and the specific heat of water is 4186 J/kg ∘C.

Answer 1

heat lost by iron = heat gained by aluminum + water

or, m_is_i(210^∘C - 28.5^∘C) = m_ws_w(28.5^∘C - 20^∘C) + m_as_a(28.5^∘C - 20^∘C)

or, m_i(450 J/kg ∘C)(210^∘C - 28.5^∘C) = (1 kg)(4186 J/kg ∘C)(28.5^∘C - 20^∘C) + (0.2 kg)(920 J/kg ∘C)(28.5^∘C - 20^∘C)

or, m_i = 0.45 kg - mass of the iron piece

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