Answer:
417 °C
Explanation:
From the question given above, the following data were obtained:
Mass of iron (Mᵢ) = 645 g
Specific heat capacity of iron (Cᵢ) = 0.449 J/gºC
Mass of water (Mᵥᵥ) = 375 g
Initial temperature of water (Tᵥᵥ) = 26 °C
Equilibrium temperature (Tₑ) = 87 °C
Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC
Initial temperature of iron (Tᵢ) =?
The initial temperature of iron can be obtained as follow:
Heat lost by iron = heat gain by water
MᵢCᵢ(Tᵢ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)
645 × 0.449(Tᵢ – 87) = 375 × 4.184 (87 –26)
289.605(Tᵢ – 87) = 1569 × 61
289.605Tᵢ – 25195.635 = 95709
Collect like terms
289.605Tᵢ = 95709 + 25195.635
289.605Tᵢ = 120904.635
Divide both side by 289.605
Tᵢ = 120904.635 / 289.605
Tᵢ ≈ 417 °C
Thus, the original temperature of the iron is 417 °C
