Problems dealing with calorimetry

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SUMMARY

The discussion focuses on solving a calorimetry problem involving a 3.0 kg gold bar at 99 degrees Celsius and 0.22 kg of water at 25 degrees Celsius. The correct equation to use is the conservation of energy, expressed as cx,mx,(Tf-Tix) = cy,my,(Tf-Tiy). Participants clarified that the change in heat (Q) of one system is equal to the negative change in heat of the other system, ensuring energy conservation. The user resolved their issue by correcting a calculation error.

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Hey, I am working on some problems dealing with calorimetry. The problem I am stuck on is: What is the final temperature when a 3.0 kg gold bar at 99 degrees C is dropped into a 0.22 kg of water at 25 degrees C.

The equation I am working with is: cx,mx,(Tf-Tix) = cy,my,(Tf-Tiy)
It seems like it should be simple to solve, but I keep getting the wrong answer. Any help?
 
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Your equation should be a conservation of energy.

Change in Q of one system plus the change in Q of another system has to be equal to 0.

What you said is change in Q of one system is equal to change in Q of the other system. This is not true. That would mean that if one system lost 100 Joules of energy so would the other one. In fact if one loses 100 Joules then the other one gains 100 Joules.
 
Sorry, that was a typo, it should read like you said.

Okay, now I got it. I think I punched it in wrong with the calculator. Thanks!
 

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