Finding equilibrium temperature of water and a solid

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Homework Help Overview

The problem involves determining the final equilibrium temperature when a mass of glass at a high temperature is added to a smaller mass of water at a lower temperature. The specific heat capacities and mass of both substances are provided, and the context suggests a focus on heat transfer and phase changes.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the heat transfer equations and the calculations leading to an apparent discrepancy between the calculated and expected equilibrium temperature. Questions are raised about the behavior of water at its boiling point and the implications of heat energy beyond that temperature.

Discussion Status

The discussion is ongoing, with participants exploring the implications of water reaching its boiling point and the potential for vaporization. Some guidance has been offered regarding the behavior of water at high temperatures, but there is no explicit consensus on the correct interpretation of the problem.

Contextual Notes

There is a mention of an answer key providing a different final temperature, which raises questions about the assumptions made in the calculations, particularly regarding phase changes and the state of water at high temperatures.

Julianj
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Homework Statement



If 100 grams of glass (specific heat .2 cal/g*°C) at a temperature of 500 °C is added to 20 g of 0°C water, what is the final equilibrium temperature?

Homework Equations



Q= mcΔT


The Attempt at a Solution



I know that I have to set the two equation against each other:

(100 g)*(.2 cal/g*°C)*(500°C -T) = (20 g)*(0°C+T)

20*(500-T)= 20T
10,000 - 20T= 20T
10,000 = 40T
250 = T

I have an answer of 250°C, but my answer key says that the correct answer is 100°C. Any help?? Thank you!
 
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Julianj said:

Homework Statement



If 100 grams of glass (specific heat .2 cal/g*°C) at a temperature of 500 °C is added to 20 g of 0°C water, what is the final equilibrium temperature?

Homework Equations



Q= mcΔT


The Attempt at a Solution



I know that I have to set the two equation against each other:

(100 g)*(.2 cal/g*°C)*(500°C -T) = (20 g)*(0°C+T)

20*(500-T)= 20T
10,000 - 20T= 20T
10,000 = 40T
250 = T

I have an answer of 250°C, but my answer key says that the correct answer is 100°C. Any help?? Thank you!

What happens when water reaches 100 C and there's still heat energy available?
 
Ah, it vaporizes! So whenever the temperature rises enough to go past 100°C, the equilibrium will be at 100, because the rest of the water vaporizes, correct? Thank you! It is this sort of thing that causes me to miss points on our exams.
 
Julianj said:
Ah, it vaporizes! So whenever the temperature rises enough to go past 100°C, the equilibrium will be at 100, because the rest of the water vaporizes, correct?
Well, at least some of the water will evaporate. You could end up with a mixture of liquid water and steam at 100C. If there's enough heat available to evaporate all of the water then the steam temperature could also rise. You should check whether or not any water will remain liquid.
Thank you! It is this sort of thing that causes me to miss points on our exams.
Glad to help :smile:
 

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