How Much Ice Was Added to Cool the Lemonade?

AI Thread Summary
To determine how much ice was added to a 0.75 kg jug of lemonade cooling from 15°C to 8°C, the heat transfer equation Q = cm∆T is utilized. The specific heat capacity of water is assumed for the lemonade, and the mass of the jug is not needed since the final temperature refers to the mixture. The discussion highlights the necessity of considering the latent heat of fusion for the ice, leading to the equation Q = cm∆T + mLv. The heat lost by the lemonade equals the heat gained by the ice, integrating both temperature change and phase change. Understanding these principles is crucial for solving the problem accurately.
chops369
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Homework Statement


How much ice was added to a 0.75 kg jug of lemonade at 15 degrees celcius if the final temperature of the jug is 8 degrees celcius? (Assume the lemonade has the specific heat capacity of water.)


Homework Equations


Q = cm∆T


The Attempt at a Solution


Well I think I might have to rearrange Q = cm∆T to solve for mass, so m = Q/c*∆T. But I don't have the heat, and then how would I account for the mass of the jug? Help! :bugeye:
 
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You don't need the mass of the jug - it means the final temperautre of the mixture in the jug.
Remember the energy needed to melt ice.
 
So you're saying I need to use latent heat? If that's the case then I would use Q = cm∆T + mLv, right?
 
chops369 said:
So you're saying I need to use latent heat? If that's the case then I would use Q = cm∆T + mLv, right?
Yes, and that equals the Q = cm∆T lost by the warm lemonade to get to the final temperature.
 

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