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

Physics
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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 20th, 2015

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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 = 23380 Joules

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 = 7.2˚
The water’s temperature decreased 7.2˚ as all the ice melted. Final temperature of water = 28 - 7.2 = 20.7˚
Now you have 770grams of water at 20.7˚C and 70 grams of water at 0˚

3rd determine the final temperature of the water.
  70* 4.186 * (Tf – 0) = 770 * 4.186 * (20.7 – Tf)
Divide both sides by 4.186
  70* Tf = 770 * (20.7 – Tf)

Tf = 18.975˚

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

Heat absorbed by ice + heat absorbed by 0˚water = Heat released by 28˚ water


Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 20th, 2015

70 * 334 + 70 * 4.186 * (Tf – 0) = 770 * 4.186 * (28 – Tf)
23380+ 293.02 Tf = 90250.16 – 3223.22 Tf
  3516.24* Tf = 66870.16
Tf = 19.02˚C

Jun 20th, 2015

please give me good reputation

Jun 20th, 2015

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