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A 25.0 g block of copper (specific heat capacity 0.380 j/g･°c) at 88.0 °c is placed into 500.0 g of water initially at 20.0 °c. what is the change in temperature (in °c) of the copper block? (the specific heat capacity of water is 4.184 j/g･°c).

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Eduard22sly

The change in temperature (in °c) of the copper block placed into the water is 67.7 °C

How to determine the equilibrium temperature

  • Mass copper (M꜀) = 25 g
  • Specific heat capacity of copper (C꜀) = 0.380 J/gºC
  • Temperature of copper (T꜀) = 88 °C
  • Mass of water (Mᵥᵥ) = 500 g
  • Temperature of water (Tᵥᵥ) = 20 °C
  • Specific heat capacity of the water = 4.184 J/gºC
  • Equilibrium temperature (Tₑ) =?

Heat loss = Heat gain

M꜀C(T꜀ – Tₑ) = MᵥᵥC(Tₑ – Tᵥᵥ)

25 × 0.38 (88 – Tₑ) = 500 × 4.184(Tₑ – 20)

9.5(88 – Tₑ) = 2092(Tₑ – 20)

Clear bracket

836 – 9.5Tₑ = 2092Tₑ – 41840

Collect like terms

836 + 41840 = 2092Tₑ + 9.5Tₑ

42676 = 2101.5Tₑ

Divide both side by 2101.5

Tₑ = 42676 / 2101.5

Tₑ = 20.3 °C

How to determine the change in temperature of copper

  • Temperature of copper (T꜀) = 88 °C
  • Equilibrium temperature (Tₑ) = 20.3 °C
  • Change in temperature (ΔT) =?

ΔT = T꜀ – Tₑ

ΔT = 88 – 20.3

ΔT = 67.7 °C

Learn more about heat transfer:

https://brainly.com/question/6363778

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