Ice Cubes & Liquid Water - Thermodynamics

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Discussion Overview

The discussion revolves around a thermodynamics problem involving the mixing of ice cubes and liquid water, specifically focusing on the final temperature of the mixture when no heat is lost to the surroundings. The context is a homework-related inquiry into heat transfer principles.

Discussion Character

  • Homework-related, Technical explanation

Main Points Raised

  • The initial setup involves four ice cubes at 0 °C and water at 75 °C, with a total mass of 53.5 g for the ice and 115 g for the water.
  • One participant states the heat gained by the ice cubes equals the heat lost by the water, presenting a basic thermodynamic relationship.
  • A participant questions the feasibility of ending up with water and ice at 21 °C, implying a misunderstanding of the phase change involved.
  • Another participant acknowledges a mistake regarding the latent heat, indicating a need to consider this factor in the calculations.
  • Expressions of gratitude and a willingness to re-attempt the problem are noted, reflecting a collaborative approach to resolving the misunderstanding.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct final temperature, and there is a clear disagreement regarding the treatment of latent heat in the calculations.

Contextual Notes

Limitations include the omission of specific formulas and the need to account for latent heat in the heat transfer calculations, which remains unresolved in the discussion.

ahaanomegas
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Homework Statement



Four ice cubes at exactly 0 \ { }^{\circ} \text{C} with a total mass of 53.5 \ \text{g} are combined with 115 \ \text{g} of water at 75 \ { }^{\circ} \text{C} in an insulated container. If no heat is lost to the surroundings, what will be the final temperature of the mixture?

Homework Equations



No specific formulas. Just basic thermodynamics, as explained in the solution.

The Attempt at a Solution



Magnitude of heat gained by Ice cubes = Magnitude of heat lost by Water
4m_{\text{one ice cube}}c_{\text{ice}}(T'-T_{\text{0-ice}})=m_{\text{water}}c_{\text{water}}(T_{\text{0-water}}-T')
From here, I used the numbers in my book and got the wrong answer [below]. :(
21 Degrees Celsius
 
Last edited:
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You mean you end with water and ice at 21°C?
 
Yes, Sir, but that's my answer, not the correct one.
 
Sigh.

Have you ever seen ice at 21°C?
 
Ah! I guess I forgot the latent heat. That was obliviously silly. Thanks for helping me through my silliness! I'll re-try the problem once I can and, hopefully, I'll get it right this time around. :) Sorry and thank you, again.
 

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