What is the equilibrium temperature of copper and water?

AI Thread Summary
The discussion revolves around calculating the equilibrium temperature when a 50 kg block of copper at 80°C is placed in 120L of water at 25°C. The initial calculations using the heat lost by copper equating to the heat gained by water led to an incorrect result below 25°C. Participants emphasize the importance of showing the working steps to identify errors in the calculations. A corrected approach suggests using the total heat change equation set to zero for clarity. The conversation highlights the significance of proper equation setup in thermodynamic problems.
Altairs
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I am not a novice at thermodynamics but still the following question made me think over my status more than twice.

I have to find the equilibrium temperature when a 50 Kg block of copper at 80C is placed in a container which as 120L of water at 25C.

I made simple equations of heat lost = heat gained. Used Specific Heat for water = 4.18 and for copper 0.285 (and 0.391) and used mcdelT. But the answer I get is below 25C. Where as it should be between 25 and 80.
 
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Altairs said:
I made simple equations of heat lost = heat gained. Used Specific Heat for water = 4.18 and for copper 0.285 (and 0.391) and used mcdelT. But the answer I get is below 25C. Where as it should be between 25 and 80.

Hi Altairs! :smile:

You know you should show us your working … otherwise how can we see where you went wrong? :confused:
 
Oh I forgot the PF SOP :rolleyes:

Heat Lost (by copper) = Heat Gained (by water)

50 * 0.385 (T_{e} - 80) = 0.12 * 1000 *4.18 (T_{e} - 25)
 
Altairs said:
Oh I forgot the PF SOP :rolleyes:

Heat Lost (by copper) = Heat Gained (by water)

50 * 0.385 (T_{e} - 80) = 0.12 * 1000 *4.18 (T_{e} - 25)

hmm … that's why we ask to see your working!

no wonder they both got colder! :rolleyes:

It would be less confusing if you said "total heat change = 0, therefore:"

50 * 0.385 (T_{e} - 80)\,+\,0.12 * 1000 *4.18 (T_{e} - 25)\,=\,0 :smile:
 
Ooh ! Thanks :blushing:
 
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