Phase Changes and Energy Conservation

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Homework Help Overview

The discussion revolves around a problem involving phase changes and energy conservation, specifically the interaction between lemonade and an ice cube. The scenario includes calculating the final temperature of the system and determining the amount of ice remaining after heat exchange.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculations of heat energy for both the lemonade and the ice cube, questioning the assumptions made about energy transfer and the melting of ice. There are attempts to clarify the calculations regarding the heat required to melt the ice and the amount of ice that can be melted with the available heat.

Discussion Status

Participants are actively engaging with the problem, offering calculations and questioning each other's reasoning. Some have provided insights into the heat exchange process and the implications of the calculations, but there is no explicit consensus on the final outcomes yet.

Contextual Notes

There are discussions about the heat capacities involved and the assumptions regarding the system being isolated from external heat exchange. Participants are also clarifying the definitions and values used in their calculations.

mikep
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A large punch bowl holds 4.75 kg of lemonade (which is essentially water) at 20.0°C. A 2.00 kg ice cube at -10.2°C is placed in the lemonade. What is the final temperature of the system, and the amount of ice (if any) remaining? Ignore any heat exchange with the bowl or the surroundings.

can someone please tell me how to solve this problem?
so far i have:
(4.75kg)(4186J/kg°C)(20°C) = 397670J
(2kg)(2090J/kg°C)(10.2°C) = 426360J
(2kg)(335000J/kg) = 670000J
 
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mikep said:
A large punch bowl holds 4.75 kg of lemonade (which is essentially water) at 20.0°C. A 2.00 kg ice cube at -10.2°C is placed in the lemonade. What is the final temperature of the system, and the amount of ice (if any) remaining? Ignore any heat exchange with the bowl or the surroundings.

can someone please tell me how to solve this problem?
so far i have:
(4.75kg)(4186J/kg°C)(20°C) = 397670J
(2kg)(2090J/kg°C)(10.2°C) = 426360J


(2kg)(2090J/kg°C)(10.2°C) = 42636J


You see the heat energy of water is enough to rise the temperature of all the ice up to 0.0 °C. The remaining 355034 J is enough to melt 335034/335000=1.06 kg ice only. I hope you can proceed from here.

ehild
 
wouldn't it be 335034/670000 = 0.53kg ? because Q = m L = (2kg)(335000) = 670000
 
mikep said:
wouldn't it be 335034/670000 = 0.53kg ? because Q = m L = (2kg)(335000) = 670000

L= 335000 J/kg. It is the heat needed to melt 1 kg ice at 0 °C. You have Q = 355034 J available heat energy from the warm lemonade.
m=Q/L=355034 J/335000 (J/kg)=1.06 kg.

That 670000 J you suggested is the heat needed to melt all the 2kg of ice. The heat available is enough to melt 0.53 times the original amount , that is 2*0.53=1.06 kg.

ehild
 
oh i get it. so te amount of heat left is 0.94kg and the final temperature would be 0°C because there is still some ice left
 
mikep said:
oh i get it. so te amount of heat left is 0.94kg and the final temperature would be 0°C because there is still some ice left

This is all right if you meant 0.94 kg ice instead of heat :smile:

ehild
 

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