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)

Answer 1

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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