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The specific heat of copper is 0.385 J/gºC. A 95.0 g block of
copper is heated to 77.9°C and placed into a calorimeter that
a
contains 42.0 mL of water at 20.4°C. Assuming no heat loss,
what will be the final temperature of the copper and water?

Sagot :

Eduard22sly

The final temperature of the copper and water in the calorimeter given the data is 30.3 °C

Data obtained from the question

  • Specific heat capacity of copper (C꜀) = 0.385 J/gºC
  • Mass of copper (M꜀) = 95 g
  • Temperature of copper (T꜀) = 77.9 °C
  • Mass of water (Mᵥᵥ) = 42 mL = 1 × 42 = 42 g
  • Temperature of water (Tᵥᵥ) = 20.4 °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

MᵥᵥC(Tᵥᵥ – Tₑ) = M꜀C(Tₑ – M꜀)

95 × 0.385 (77.9 – Tₑ) = 42 × 4.184(Tₑ – 20.4)

36.575(77.9 – Tₑ) = 175.728(Tₑ – 20.4)

Clear bracket

2849.1925 – 36.575Tₑ = 175.728Tₑ – 3584.8512

Collect like terms

2849.1925 + 3584.8512 = 175.728Tₑ + 36.575Tₑ

6434.0437 = 212.303Tₑ

Divide both side by 212.303

Tₑ = 6434.0437 / 212.303

Tₑ = 30.3 °C

Learn more about heat transfer:

https://brainly.com/question/6363778

#SPJ1

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