What is the equilibrium temperature of copper and water?

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SUMMARY

The equilibrium temperature of a 50 kg block of copper at 80°C placed in 120 liters of water at 25°C can be calculated using the principle of conservation of energy, where heat lost by copper equals heat gained by water. The specific heat capacities used are 0.385 J/g°C for copper and 4.18 J/g°C for water. The correct equation to determine the equilibrium temperature (T_e) is 50 * 0.385 (T_e - 80) + 0.12 * 1000 * 4.18 (T_e - 25) = 0. This ensures the final temperature lies between 25°C and 80°C, resolving the initial confusion regarding the temperature being below 25°C.

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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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