How Much Ice Was Added to Cool the Lemonade?

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Homework Help Overview

The problem involves determining the mass of ice added to a jug of lemonade to achieve a specific final temperature. The context is centered around thermal energy transfer and the specific heat capacity of water.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss rearranging the heat equation to isolate mass and express concerns about missing information, such as the heat involved. There is also a mention of the need to consider the energy required to melt ice and how to incorporate latent heat into the calculations.

Discussion Status

Some participants have provided guidance on the relevance of latent heat in the calculations, suggesting a relationship between the heat lost by the lemonade and the heat gained by the ice. Multiple interpretations of the problem are being explored, particularly regarding the equations to use.

Contextual Notes

There is a discussion about whether the mass of the jug should be considered, with some participants indicating it is not necessary for the final temperature calculation.

chops369
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Homework Statement


How much ice was added to a 0.75 kg jug of lemonade at 15 degrees celsius if the final temperature of the jug is 8 degrees celsius? (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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