##### A(n) 90-g ice cube at 0°C is placed in 670 g of water at 24°C. What is the final

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A(n) 90-g ice cube at 0°C is placed in 670 g of water at 24°C. What is the final temperature of the mixture?

°C

Oct 17th, 2017

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Jun 19th, 2015

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 x Heat of fusion = 90 x 334 = 30,060 Joules

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

Energy released by water = Mass x Specific heat x ∆T
Energy released by water = 670 x 4.18 x ∆T

670 x 4.186 x ∆T = 90 x 334
∆T = 10.7˚
The water’s temperature decreased 10.7˚ as all the ice melted. Final temperature of water = 24 - 10.7 = 13.3˚
Now you have 670 grams of water at 13.3˚C and 90 grams of water at 0˚

3rd determine the final temperature of the water.
90 x 4.186 x (Tf – 0) = 670 x 4.186 x (13.3 – Tf)
Divide both sides by 4.186
90 x Tf = 670 x (13.3 – Tf)
760 x Tf = 8911
Tf = 11.7˚

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 24˚ water

90 x 334 + 90 x 4.186 x (Tf – 0) = 670 x 4.186 x (24 – Tf)
30060 + 376.2 Tf = 67214.4 – 2800.6 Tf
3176.8 x Tf = 37154.4
Tf = 11.7˚C

Jun 19th, 2015

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