Answered

Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

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)

Sagot :

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

Read more on heat:

https://brainly.com/question/21406849

We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.