Respuesta :
Heat gained by ice cubes would be equal to the - heat lost by warm water
The moles of ice is: 50.5 g / 18.0 g/mol = 2.81 mol
Heat required to melt all of the ice is equal to: 2.81 mol X 6.02 kJ/mol = 16.9 kJ = 16890 J
Now, know whether the warm water will still be above 0C when it loses this much heat:
-1690 J = 160 g (4.184 J/gC) (Delta T) Delta T = -25C
In order to solve for the final temperature, going back to include warming of the melted ice to a final temperature:
q(ice/water) = - q(warm water)
moles (Delta Hf) + m c (T2-T1) = - m c (T2-T1)
50.5 g / 18.0 g/mol = 2.81 mol
2.81 mol X 6.02 kJ/mol + 50.5g (4.184 J/gC) (T2-0) = -160g (4.184 J/gC) ( T2-80)
16916 + 211.3T2 = -669.4 T2 + 53555
36639 = 880.7 T2
T2 = 41.6 C
The moles of ice is: 50.5 g / 18.0 g/mol = 2.81 mol
Heat required to melt all of the ice is equal to: 2.81 mol X 6.02 kJ/mol = 16.9 kJ = 16890 J
Now, know whether the warm water will still be above 0C when it loses this much heat:
-1690 J = 160 g (4.184 J/gC) (Delta T) Delta T = -25C
In order to solve for the final temperature, going back to include warming of the melted ice to a final temperature:
q(ice/water) = - q(warm water)
moles (Delta Hf) + m c (T2-T1) = - m c (T2-T1)
50.5 g / 18.0 g/mol = 2.81 mol
2.81 mol X 6.02 kJ/mol + 50.5g (4.184 J/gC) (T2-0) = -160g (4.184 J/gC) ( T2-80)
16916 + 211.3T2 = -669.4 T2 + 53555
36639 = 880.7 T2
T2 = 41.6 C
Based on the data provided, the final temperature of the mixture is 49.1°C.
What is the final temperature of the mixture?
Based on the principle of conservation of energy:
- Heat gained by ice cubes would be equal to the heat lost by warm water
- Also, Q = mc◇T
- Also Q = mL
moles of ice cube = mass/molar mass
molar mass of ice/water = 18m0 g/mol
moles of ice = 50.5 g / 18.0 g/mol = 2.81 moles
Heat required to melt all of the ice H = 2.81 mol X 6.02 kJ/mol
Heat required = 16.9 kJ = 16890 J.
Heat required to convert water at 0°C to a final temperature T2, H = mc◇T
Where ◇T = T2 - T1
H = 50.5 × 4.18 × (T2 - 0)
H = 211.09T
Total heat gained = 16980 + 211.09T
Heat lost by water at 90°C = m c (T1 - T2)
H = -160g (4.184 J/gC) ( 90 -T2)
H = 60192 - 668.8T2
Using heat gained = heat lost
16980 + 211.09T = 60192 - 668.8T2
60192 - 16980 = 668.8T2 + 211.09T2
43212 = 879.89T2
T2 = 49.1°C
Therefore, the final temperature of the mixture is 49.1°C.
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T2 = 41.6 C