Calculating Final Temperature and State When Mixing Ice and Water

AI Thread Summary
When mixing 200g of ice at -5°C with 20g of water at 15°C, the final state will be a block of ice at a temperature between -5°C and 0°C due to the larger mass of ice. The heat gained by the ice must equal the heat lost by the water, but the phase change complicates the calculation. The relevant equations include Q=mcΔT for temperature changes and Q=mLf for phase changes. The discussion emphasizes the need to account for the phase change of ice melting into water. The final equilibrium state will depend on the balance of heat transfer between the ice and water.
DBW3
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Homework Statement



If you mixed 200g if ice that is at -5°C with 20g of water that is at 15°C, what will be the temperature and condition of the final state once equilibrium is achieved?


Homework Equations


Q=mcΔT, Q=mLf


The Attempt at a Solution


I know that the amount of ice is to great to change with such little water, so I know the final condition will be a 220g block of ice at some temperature between -5°C and 0°C. Since the mass of the water is added to the mass of the ice does this imply that Qgained≠Qlost?

That's what I'm having trouble on, deriving the equation to solve for temperature if they are not equal.
 
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Go ahead with your equations and whatever you know.i shall help you along.
 
Hi DBW, have you considered the phase change involved? :smile:
 
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