Question

A 645 g piece of iron is plunged into 375 g of water. The temperature of the water increases from 26 C to 87 C. If the heat capacity of iron is 0.449 J/g C, what must have been the original temperature (in C) of the piece of iron

Answer 1

417 °C

Step-by-step 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