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
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, = 28⁰Cmass of the ice, = 5.0 gtemperature of the ice, = 0⁰ Cspecific heat capacity of ice = 2.09 J/g ⁰Cheat of fusion of ice = 333.55 J/gspecific heat capacity of water = 4.184 J/g⁰ Cdensity of water = 1 g/mlLet 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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