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Thermal Properties of Solids and Gases Problem

  • Thread starter Silverbolt
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  • #1
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At a local county fair, you watch as blacksmith drops a 0.050-kg iron horseshoe into a bucket containing 25.0 kg of water. If the initial temperature of the horseshoe is 450°C, and the initial temperature of the water is 23°C, what is the equilibrium temperature of the system?


Here is how i am doing it (PLEASE CORRECT ME IF I AM WRONG)

mcΔT= mcΔT
(.050 kg)(448 J/kg(°C) )(Tf-450°C) = (25.0 kg)(4186 J/kg(°C) )(Tf-23°C)
 

Answers and Replies

  • #2
cepheid
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It seems perfectly fine to me so far, up to a negative sign, at least.

You should make sure you understand what physics is happening. The heat (Q) flowing out of the iron goes into the water. They have opposite directions (one loses energy in the heat transfer, while the other gains it), so there should be a difference in sign:

QFe = -QH2O

mFecFeΔTFe = -mH2OcH2OΔTH2O

That's the physics. The rest is all computation.

EDIT: you can see that you need the negative sign there just for arithmetic purposes, since the temperature change of the iron is negative, while the temperature change of the water is positive.
 
  • #3
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So if I include the negative sign everything else will be correct?
 
  • #4
gneill
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Your approach is okay, but watch out for the signs of the ΔT's. The order of the temperatures, Tf - x versus x - Tf, matters if you want to compare positive values. Do you expect your final temperature to be higher than 450C?
 
  • #5
cepheid
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So if I include the negative sign everything else will be correct?
Well, I haven't looked to make sure your numbers are correct for the heat capacities, but yes, your method is fine.
 
  • #7
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Please help. When a ring heated why it expands only outward? why not inward also making the hole smaller.
Thank you.
 

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