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(Solved): JUST NEED ANSWER FOR Q10 - Q16 PLEASE You mix:300.0 grams of water at 25.0 oC50.0 grams of Ice at ...



JUST NEED ANSWER FOR Q10 - Q16 PLEASE

You mix:

  • 300.0 grams of water at 25.0 oC
  • 50.0 grams of Ice at –25.0 oC
  • 50.0 grams of Steam at 125.0 oC.
    • Specific Heat of Ice = 2.09 J/(goC) = 37.65 J/(moloC)
    • Specific Heat of Water = 4.184 J/(goC) = 75.37 J/(moloC)
    • Specific Heat of Steam = 2.01 J/(goC) = 36.21 J/(moloC)
    • Enthalpy of Fusion for Water = 6.02 kJ/mol
    • Enthalpy of Vaporization for Water = 40.7 kJ/mol

Let's first get each phase to Liquid water at 0 oC. We can then put energy back in to bring it to the Final Temp.

1. Will the energy to warm the Ice be Positive or Negative?

Group of answer choices

Positive since it is absorbing heat

Negative since it releases heat

2. Calculate the energy (in kJ) to warm the Ice to 0 oC. (3s.f.)

3. Will the energy to melt the Ice be Positive or Negative?

Group of answer choices

Positive since it is absorbing heat

Negative since it releases heat

It will not have a sign since it melts at a constant temperature

4. Calculate the energy (in kJ) to melt the Ice at constant 0 oC. (3s.f.)

5. Will the energy to cool the steam to 100 oC be Positive or Negative?

Group of answer choices

Positive since it is absorbing heat

Negative since it releases heat

6. Calculate the energy (in kJ) to cool the steam to 100 oC. (3s.f.)

7. Will the energy to condense the steam (to liquid water) at 100 oC be Positive or Negative?

Group of answer choices

Positive since it is absorbing heat

Negative since it releases heat

It will not have a sign since it condenses at a constant temperature

8. Calculate the energy (in kJ) to condense the steam (to liquid water) at constant 100 oC. (3s.f.)

9. Will the energy to cool the Liquid "steam" water to 0 oC be Positive or Negative?

Group of answer choices

Positive since it is absorbing heat

Negative since it releases heat

10. Calculate the energy (in kJ) to cool the liquid "steam" water from 100 oC to 0 oC. (3s.f.)

11. Will the energy to cool the original water to 0 oC be Positive or Negative?

Group of answer choices

Positive since it is absorbing heat

Negative since it releases heat

12. Calculate the energy (in kJ) to cool the original 300.0g of water from 25.0 oC to 0 oC. (3s.f.)

13. If you add the 6 energies calculated above, this will be negative and represent extra heat released if you brought everything to the liquid state at 0 oC.

What is this energy (in kJ) as a positive number (since we will be adding back into the system to reach the real Final Temperature).

14. Calculate the Final Temperature (in oC) for the 400.0 grams of water. (3s.f.)

15. Why can we use the Energy steps above even though some of the steps do not really happen (e.g. original water does not really cool to 0 oC)

Group of answer choices

All energies are State functions so we can use any pathway

Enthalpy (qp) is a State function so we can use any pathway

Ethalpy is a Path function so we can use this pathway

All energies are path functions so we can use any pathway

16. Why is the Enthalpy of Vaporization so much greater than than the Enthalpy of Fusion?

Group of answer choices

Solid-to-Liquid phase change requires more energy than Liquid-to-gas

Gas phase particles have less kinetic energy than liquid particles

Separating particles from liquid to gas phase has to fully breaks the intermolecular forces



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