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 21st, 2015

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Jun 21st, 2015

ok

Jun 21st, 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 * Heat of fusion = 90 * 334 = 30,060 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 = 670 * 4.18 * ∆T 

670 * 4.186 * ∆T = 90 * 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 665 grams of water at 16.4˚C and 80 grams of water at 0˚ 

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


Jun 21st, 2015

thank you 

Jun 21st, 2015

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Jun 22nd, 2015

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