Iron ball Mass = 0.8 kg, Cp=484, Ti=80 Celsius edit : Cp = 448
Aluminum Bucket = 0.5 kg, Cp=900, Ti=20 Celsius
Water = 0.6 kg, Cp=4186, Ti=20 Celsius
Ice cube=0.5 Kg, Cp=2090, Lf=3.33e5, Ti=-5 Celsius
Find the final temperature of the bucket.
Delta Q = 0 in equilibrium
The Attempt at a Solution
At fist I took into consideration only the water,iron and Aluminum to find the temperature in equilibrium
Delta Q = 4186*0.6*(Tf-20)+900*0.5*(Tf-20)+448*0.8(Tf-80)=0
The final temperature I got is 26.47 Celsius
Next I calculate the amount of energy release from 0.6 kg of water = 600*4.182*26.47=66,482 J
Then I divided it by the amount of energy needed to bring this amount of water to 0 Celsius - 66,482/333=199.645 Grams of ice.
Since we have 0.5 kg of ice Tf will be equal to 0 Celsius and 0.3 Kg of the ice will remain in the bucket.
In order to proof it I wrote the following equation = Delta Q = 4186*0.6*(Tf-26.47)+2090*0.199*(0-(-5)) +0.199*3.33e5+4186*(0.199)*(Tf-0)=0
I got Tf= -0.65 Celsius which could be a result of roundoff error so it seems like I got the right answer.
But I think that I did something wrong since I ignored the Iron and the Aluminum in my last equation.
Thank you for your help!