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In an insulated cup of negligible heat capacity, 50. G of water at 40. °c is mixed with 30. G of water at 20. °c. The final temperature of the mixture is closest to.

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Eduard22sly

The final temperature of the mixture is closest to 32.5 °C

Data obtained from the question

  • Mass of warm water (Mᵥᵥ) = 50 g
  • Temperature warm water (Tᵥᵥ) = 40 °C
  • Mass of cold water (M꜀) = 30 g
  • Temperature of cold water (T꜀) = 20 °C
  • Specific heat capacity of the water = 4.184 J/gºC
  • Equilibrium temperature (Tₑ) =?

How to determine the equilibrium temperature

Heat loss = Heat gain

MᵥᵥC(Tᵥᵥ – Tₑ) = M꜀C(Tₑ – M꜀)

50 × 4.184 (40 – Tₑ) = 30 × 4.184(Tₑ – 20)

209.2(40 – Tₑ) = 125.52(Tₑ – 20)

Clear bracket

8368 – 209.2Tₑ = 125.52Tₑ – 2510.4

Collect like terms

8368 + 2510.4 = 125.52Tₑ + 209.2Tₑ

10878.4 = 334.72Tₑ

Divide both side by 334.72

Tₑ = 10878.4 / 334.72

Tₑ = 32.5 °C

Learn more about heat transfer:

https://brainly.com/question/6363778

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