Answer:
the resulting temperature is 23.37 ⁰C
Explanation:
Given;
mass of the iron, m₁ = 80 g = 0.08 kg
mass of the water, m₂ = 200 g = 0.2 kg
mass of the iron vessel, m₃ = 50 g = 0.05 kg
initial temperature of the iron, t₁ = 100 ⁰C
initial temperature of the water, t₂ = 20 ⁰C
specific heat capacity of iron, c₁ = 462 J/kg⁰C
specific heat capacity of water, c₂ = 4,200 J/kg⁰C
let the temperature of the resulting mixture = T
Apply the principle of conservation of energy;
heat lost by the hot iron = heat gained by the water
Therefore, the resulting temperature is 23.37 ⁰C
