Find Equilibrium Temp. of Copper & Water Mixture

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Homework Help Overview

The problem involves finding the equilibrium temperature of a mixture of copper and water in an insulated container. The original poster presents specific weights and temperatures for both substances, along with their specific heat capacities.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to set up an equation based on the principle of conservation of energy, but expresses uncertainty about the setup. Some participants clarify that the heat gained by water equals the heat lost by copper, suggesting a relationship between the two.

Discussion Status

Participants are engaged in clarifying the setup of the problem and confirming the equations involved. One participant provides additional context regarding the insulated nature of the system, while another expresses confusion about the initial temperature of copper, indicating a potential typographical error. The discussion appears to be moving towards a clearer understanding of the problem.

Contextual Notes

The original poster's use of imperial units is noted, and there is a mention of a typo regarding the temperature of copper, which may affect the interpretation of the problem.

biamin
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Homework Statement


If 7.3 lbs of copper at 12540 deg. F. is added to .52 ft^3 of water at 35 deg. F. in an insulated container, find the equilibrium temperature.



Homework Equations


I Think I need to use: delta Q= c*m(Temp. final - Temp. original)
c.c. = .092 BTU/lb (specific heat capacity of copper)
c.w. = 1 BTU/lb ( specific heat capacity of water)

The Attempt at a Solution


not sure how to set up the equation. any help would be appreciated.
 
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forgot to mention m = mass
 
Okay i don't know how to use imperial so i can only guide you with explanations.

your equations are correct you just need this little bit of information

the system is insulated therefore there is no net energy loss.

the heat gained by water is the heat lost by copper

i.e Q(water)=-Q(copper).

therefore if you set up that equation (with the equations you have given), you will be able to find the final temperature. taking note of both their initial temperatures.

hope i helped.
 
thanks, just sorted out the problem.
 
Wow! Copper at 12540 deg. F! What is that? Superheated copper vapor?
 
Sorry, Typo. should have been 1250 deg. F.
 

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