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If a 40.1 g piece of iron at 652 °C is dropped into a sample of 328 g of water at 32.4 °C, what will the final temperature be after thermal equilibrium is established? Assume that no heat is lost during the process. specific heat capacity of iron is 0.45 J/g C and water is 4.184 J/g C. SHOW YOUR WORK.

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Eduard22sly

The final temperature of the mixture after thermal equilibrium is reached given the data is 40.4 °C

Data obtained from the question

  • Mass of iron (M) = 40.1 g
  • Temperature of iron (T) = 652 °C
  • Specific heat capacity of iron (C) = 0.45 J/gºC
  • Mass of water (Mᵥᵥ) = 328 g
  • Temperature warm water (Tᵥᵥ) = 32.4 °C
  • Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC
  • Equilibrium temperature (Tₑ) =?

How to determine the equilibrium temperature

Heat loss = Heat gain

MC(T – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

40.1 × 0.45(652 – Tₑ) = 328 × 4.184(Tₑ – 32.4)

18.045(652 – Tₑ) = 1372.352(Tₑ – 32.5)

Clear bracket

11765.34 – 18.045Tₑ = 1372.352Tₑ – 44464.2048

Collect like terms

11765.34 + 44464.2048 = 1372.352Tₑ + 18.045Tₑ

56229.5448 = 1390.397Tₑ

Divide both side by 1390.397

Tₑ = 56229.5448 / 1390.397

Tₑ = 40.4 °C

Learn more about heat transfer:

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