A(n) 70-g ice cube at 0°C is placed in 770 g of water at 28°C. What is the final

Physics
Tutor: None Selected Time limit: 1 Day

A(n) 70-g ice cube at 0°C is placed in 770 g of water at 28°C. What is the final temperature of the mixture? 

Jun 19th, 2015

Thank you for the opportunity to help you with your question!

Heat of fusion of ice = 334 J/g
Specific heat of water = 4.186 J/g*˚C

1st determine the amount of heat energy required to melt all the ice.
The water’s temperature will decrease as the ice melts.
Energy absorbed by ice = mass * Heat of fusion = 70 * 334 = 23,380Joules

2nd determine the temperature of the water, after releasing the heat energy.

Energy released by water = Mass * Specific heat * ∆T
Energy released by water = 770 * 4.18 * ∆T

  770* 4.186 * ∆T = 70 * 334
∆T = 0.138˚
The water’s temperature decreased 0.138˚ as all the ice melted. Final temperature of water = 28 - 9.6 = 27.862˚
Now you have 770 grams of water at 27.862˚C and 70 grams of water at 0˚

3rd determine the final temperature of the water.
  70* 4.186 * (Tf – 0) = 770 * 4.186 * (27.862 – Tf)
Divide both sides by 4.186
  70* Tf = 770 * (27.862 – Tf)

Tf = 25.54˚

Now, let’s do the entire problem at once to check the answer!


Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 19th, 2015

Did you know? You can earn $20 for every friend you invite to Studypool!
Click here to
Refer a Friend
...
Jun 19th, 2015
...
Jun 19th, 2015
Dec 3rd, 2016
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer