What is the final temperature if only one ice cube is used?

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Homework Help Overview

The discussion revolves around a thermal equilibrium problem involving ice and water. The original poster is trying to determine the final temperature when one ice cube is introduced into a specific amount of water, with initial temperatures provided for both substances.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the energy transfer needed to melt the ice and the subsequent temperature changes. There are questions about the calculations involved in determining the energy required to heat the ice and melt it, as well as how to compute the final temperature of the system.

Discussion Status

Some participants have provided guidance on calculating the energy needed for melting the ice and heating it to 0 degrees Celsius. There are indications of confusion regarding the calculations, but the discussion has included attempts to clarify the steps involved. The original poster expresses uncertainty about the process, while others are working through the math.

Contextual Notes

Participants note the absence of specific formulas from the teacher, which has contributed to the original poster's confusion. There is also mention of needing to account for the specific heat of ice and the heat of fusion for accurate calculations.

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Two 60 g ice cubes are dropped into 270 g of water in a thermally insulated container. If the water is initially at 25°C, and the ice comes directly from a freezer at -15°C

What is the final temperature if only one ice cube is used?

I have no idea where to begin...my teacher didn't give us the formulas that we would need. Any help would be appreciated.
 
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can i get some help?
 
Calculate the amount of energy you need to completely melt the ice, which will be the mass of ice times the heat of fusion. This is how much energy goes from the water to the ice. This energy loss corresponds to a drop in temperature. You will get a new temperature for the amount of water and your ice will now be water at 0 degrees (Centigrade). Computing the final temperature is a matter of taking a weighted average of the water temperatures.

You will need the heat of fusion for ice (J/g) and the specific energy (J/g*C) for water, which are constants that can easily be looked up.
 
i am sorry, I'm not following what your saying.
are you saying.. 334(60)=20040..for the first part?
 
Sorry, I forgot to mention that the ice needs to be heated to 0 degrees Centigrade before it is melted. You will need to find the specific heat of ice as well. And yes, you have calculated the energy required to melt the ice correctly, add to that the energy needed to reach 0 degrees and subtract that from the energy of the water at 25 degree Centigrade..
 
Last edited:
so.i got ...20040+450-6750=13740
 
is that correct?
 
never mind..i got the answer..thank you
 

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