Ice cube placed in water - find final temperature

AI Thread Summary
To find the final temperature of a mixture of ice and water, the heat transfer equation must account for both the melting of the ice and the temperature change of the water. The initial setup includes a 75-gram ice cube at 0 degrees Celsius and 825 grams of water at 25 degrees Celsius. The user attempted to solve the problem using specific heat capacities but mixed units, which could lead to incorrect results. Additionally, the latent heat of fusion for the ice must be considered, as it needs to melt before the temperature of the mixture can be calculated. The expected final temperature of the mixture is 16 degrees Celsius, highlighting the importance of correctly applying thermodynamic principles.
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Homework Statement


A 75 gram ice cube at 0 degrees Celcius is placed in 825 grams of water at 25 degrees C. What is the final temperature of the mixture?


Homework Equations


What is the best formula to use in this question?


The Attempt at a Solution

I have attempted this problem multiple times. The last time I plugged in these variables:
(75g)(2.108kJ/kg-k)(T2-0)=(825g)(4.184J/g*degrees C)(T2-25 degrees C)
The answer is supposed to be 16 degrees C. I can't seem to figure this one out. Any help will be greatly appreciated. Am I missing variables?
 
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Well, I can think of two things you might check on the left side of your equation you seem to have mixed units you have 75g of ice but write J/kg.

Also, don't you have to melt the ice and wouldn't that involve latent heat?
 

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