A 85-g ice cube at 0°C is placed in 785 g of water at 28°C. What is the final temperature of the mixture
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A 85-g ice cube at 0°C is placed in 785 g of water at 28°C. What is the final temperature of the mixture

[From: ] [author: ] [Date: 11-04-29] [Hit: ]
6800cal + 85 (10)* (t2 - 0) = 785g (1)(T1water-T2) where the final temperature of the ice is equal to the final temperature of the cooled water. T1=28, Cwater=1 and t2= the final temperature.Solve for t2 = 17.......
Hi, if someone could help me set this up with an equation I would be delighted. I have tried to use q=mc/delta T, and I don't know if this is correct, because I am getting the wrong answer. I'm not asking for the answers so much as I'd like to know how to set it up. Thank you for you time.

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The correct equation is Q=mCDeltaT

Calculate the heat required to melt the ice to water at 0 Deg C = 85 gm x 80 cal/gm = 6800 cal

The heat gained by the ice = heat lost by 785 g of water

6800cal + 85 (10)* (t2 - 0) = 785g (1)(T1water-T2) where the final temperature of the ice is equal to the final temperature of the cooled water. T1=28, Cwater=1 and t2= the final temperature.

Solve for t2 = 17.45 deg C
1
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