Thermal Properties of Solids and Gases Problem

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SUMMARY

The discussion focuses on calculating the equilibrium temperature of a system involving a 0.050-kg iron horseshoe at 450°C dropped into 25.0 kg of water at 23°C. The heat transfer is analyzed using the principle of conservation of energy, expressed as mcΔT for both the iron and water. The correct formulation includes a negative sign to account for the opposite directions of heat flow, ensuring accurate temperature change calculations. The participants confirm that understanding the signs of ΔT is crucial for correct arithmetic and final temperature expectations.

PREREQUISITES
  • Understanding of heat transfer principles
  • Familiarity with specific heat capacity values (e.g., 448 J/kg(°C) for iron, 4186 J/kg(°C) for water)
  • Basic algebra for solving equations
  • Knowledge of thermodynamic equilibrium concepts
NEXT STEPS
  • Review the concept of conservation of energy in thermodynamic systems
  • Learn about specific heat capacities of various materials
  • Study the effects of temperature changes on solid and liquid states
  • Explore advanced heat transfer calculations involving multiple phases
USEFUL FOR

Students studying thermodynamics, physics educators, and anyone interested in understanding heat transfer and thermal properties of materials.

Silverbolt
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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)
 
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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.
 
So if I include the negative sign everything else will be correct?
 
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?
 
Silverbolt said:
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.
 
Thank You!
 
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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