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

Physics
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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

Jun 19th, 2015

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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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