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

Question

Nutka1998 [239]

3 years ago

7

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)

Chemistry

1 answer:

Stells [14] · Answer 1 · 2022-11-24T02:40:53+03:00

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;

<em>DH = m × c × DT</em>

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