Find Equilibrium Temp. of Copper & Water Mixture

AI Thread Summary
To find the equilibrium temperature of a copper and water mixture, the heat gained by the water equals the heat lost by the copper, as the system is insulated. The specific heat capacities of copper and water are provided, with copper at 0.092 BTU/lb and water at 1 BTU/lb. The correct setup involves using the equation Q(water) = -Q(copper) to solve for the final temperature, taking into account their initial temperatures. A typo was noted regarding the copper's temperature, correcting it from 12540°F to 1250°F. This adjustment is crucial for accurate calculations.
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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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