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Calculate the final temperature (once the ice has melted) of a mixture made up initially of 70.0 mL liquid water at 28 oC and 5.0 g ice at 0.0 oC.

Respuesta :

ajeigbeibraheem

Answer:

21.52° C

Explanation:

From the given information:

mass of the liquid water = 70.0 mL

Initial temperature = 20°C

mass of the ice = 5.0 g

temperature of ice = 0.0°C

Using the calorimetric function:

heat lost by water = heat gained by ice.

mass of water × specific heat of water (s) × ΔT = mass of ice × specific heat of ice (s) × ΔT + n (ΔH_fusion}

⇒ 70 × 4.184 × (28 -x) = 5 × 2.108(x - 0) + [tex](\dfrac{5}{18})[/tex] × 6.01 × 10³

By solving the above equation,

x = 21.52° C

onyebuchinnaji

The final temperature of the mixture of water and ice is 21.5 ⁰C.

The given parameters;

  • initial volume of the liquid = 70 mL
  • initial temperature of the water, = 28⁰C
  • mass of the ice, = 5.0 g
  • temperature of the ice, = 0⁰ C
  • specific heat capacity of ice = 2.09 J/g ⁰C
  • heat of fusion of ice = 333.55 J/g
  • specific heat capacity of water = 4.184 J/g⁰ C
  • density of water = 1 g/ml

Let the final temperature of the mixture = t

mass of the liquid water = 1 g/ml  x 70 ml = 70 g

Apply the principle of conservation of energy to determine the final temperature of the mixture;

heat lost by water = heat gained by ice

[tex]70 \times 4.184(28 - t) = 5\times 2.09(t - 0) \ + \ 5 \times 333.55\\\\8200.64 - 292.88t = 10.45 t + 1667.75\\\\303.33t = 6532.89\\\\t = \frac{6532.89}{303.33}\\\\t = 21.5 \ ^0C[/tex]

Thus, the final temperature of the mixture of water and ice is 21.5 ⁰C.

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