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If 100mL of boiling water is combined with 25g of ice, what will its
temperature be when equilibrium is reached?
You’ll get brainlest but I need it rn rn

Respuesta :

Edufirst

Answer:

  • 85ºC

Explanation:

Assume that all the heat released by the boiling water is absorbed by the ice.

1. Heat released by the 100 mL liquid boiling water

  • 100 mL × 100g/mL = 100g
  • Q = mass × specific heat × ΔT
  • Q = 100g × 4.186J/g × (100ºC - T)

2. Heat absorbed by the 25 g of ice

a) Heat to melt the ice

  • Assume the initial temperature of ice is 0ºC
  • Heat to melt the ice = latent fusion heat × mass
  • L = 334J/g × 25g = 8,350J

b) Heat to increase of liquid water from 0ºC until the equilibrium temperature

  • Q = mass × specitif heat × ΔT
  • Q = 25g ×4.186J/g × (T - 25ºC)

3. Set your final equation:

  • 100g × 4.186J/g × (100ºC - T) = 25g ×4.186J/g × (T - 25ºC)

Solve:

  • 41,860 - 418.6T = 104.25T - 2,616.25
  • 523.25T = 44,476.25
  • T = 85ºC

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