40.0 g of ice cubes at 0.0°C are combined with 150. g of liquid water at 20.0°C in a coffee cup calorimeter. Calculate the final temperature reached, assuming n
Chemistry
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Question
40.0 g of ice cubes at 0.0°C are combined with 150. g of liquid water at 20.0°C in a coffee cup calorimeter. Calculate the final temperature reached, assuming no heat loss or gain from the surroundings. (Data: specific heat capacity of H2O(l), c = 4.18 J/g×°C; H2O(s) => H2O(l) DH = 6.02 kJ/mol)Calculate the final temperature reached, assuming no heat loss or gain from the surroundings. (Data: specific heat capacity of H2O(l), c = 4.18 J/g×°C; H2O(s) => H2O(l) DH = 6.02 kJ/mol)
1 Answer

1. User Answers adioabiola
The final temperature of the mixture in the coffee cup calorimeter is; 19.467 °C
According to the law of energy conservation:
As such; the heat transfer in the liquid water is equal to heat gained by the ice
Heat transfer by liquid water is therefore;
 DH = m × c × DT
 DH = 6.02 kJ/mol) = 150 × 4.18 × (T1  T2)
 6020 J/mol = 627 × (20  T2)
However, since 18g of water makes one mole
 6020 J/mol = 6020/18 = 334.44 J/g.
 334.44 = 627 × (20  T2)
 0.533 = (20  T2)
 T2 = 20  0.533
T2 = 19.467°C
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