Find Equilibrium Temp. of Copper & Water Mixture

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SUMMARY

The discussion focuses on calculating the equilibrium temperature of a copper and water mixture in an insulated container. The specific heat capacities used are 0.092 BTU/lb for copper and 1 BTU/lb for water. The key equation applied is delta Q = c * m * (Temp. final - Temp. original), where the heat gained by water equals the heat lost by copper. The final temperature can be determined by setting up the equation Q(water) = -Q(copper) and solving for the final temperature using the initial temperatures provided.

PREREQUISITES
  • Understanding of specific heat capacity (c.c. and c.w.)
  • Knowledge of heat transfer principles in insulated systems
  • Familiarity with the equation delta Q = c * m * (Temp. final - Temp. original)
  • Basic skills in unit conversion (imperial to metric)
NEXT STEPS
  • Study the principles of thermal equilibrium in insulated systems
  • Learn how to apply the conservation of energy in heat transfer problems
  • Explore specific heat capacities of various materials for comparative analysis
  • Practice unit conversions between imperial and metric systems for scientific calculations
USEFUL FOR

This discussion is beneficial for students in thermodynamics, physics enthusiasts, and anyone involved in heat transfer calculations, particularly in engineering or physical sciences.

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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