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5.00 g of ice which is at -10.0 celsius are added to 20.0 mL of hot coffee at 60.0...

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roshan-rox | Valedictorian

Posted August 13, 2013 at 6:24 PM via web

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5.00 g of ice which is at -10.0 celsius are added to 20.0 mL of hot coffee at 60.0 celsius, inside a thermos flask which is then tightly closed. What is the final temperature of the coffee?

(specific heat of ice: 2100 J/Kg K, latent heat of ice: 334 KJ/Kg)

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted August 13, 2013 at 6:36 PM (Answer #1)

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By absorbing the heat of coffee, ice will start to melt. Ice will increase its temperature until it becomes 0C ice and then using latent heat it will become 0C water. Then again it will start to increase temperature until the system comes to equilibrium. All this time coffee will decrease its temperature.

Energy required to get -10C ice to 0C `= 0.005xx2100xx10 = 105J`

Energy required to get 0C ice to 0C water `= 0.005xx334xx1000 = 1670J`

If the final temperature of the system is T;

Assume that the density of coffee is 1g/ml (same as water) and specific heat capacity of coffee is 4200J/KgK.

So mass of coffee will be 20g.

Heat loss by coffee = heat gain by ice

`0.020xx4200xx(60-T) = 105+1670+0.005xx4200xx(T-0)`

`T = 31.1C`

So the final temperature of coffee is 31.1C

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