Thermodynamics: Dropping a hot horseshoe into a pot of water

  • #1

Homework Statement


A freshly-forged iron horseshoe, with a mass of 0.549 kg is dropped into a 0.281 kg iron pot which contains 1.60 kg of water at 21.4 oC.



After the horseshoe, pot and water reach thermal equilibrium they have a temperature of 29.8 oC.

Assuming the pot and the water were in thermal equilibrium before the horseshoe entered the water, calculate the combined amount of heat gained by the pot and water as they cool the horseshoe.

From previous question:
Initial temp of horseshoe = 261.499575 degrees C
Q (water) = 56179.2 J
Q (pot) = 1062.18 J

Homework Equations


Q= mc (change in temp)


The Attempt at a Solution


Q (horseshoe) = 0.549 x 450 x (261.499575-29.8)
= 57241.38 J

Q (water) = 56179.2 J
Q (pot) = 1062.18 J
Q (pot & water) = 56179.2 + 1062.18
= 57241.38 J

Am I heading in the right direction to calculate the combined amount of heat gained by the pot and water as they cool the horseshoe???
 
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Answers and Replies

  • #2
Worked out that the answer for this question is just = 57241.38 J rounded to 3 significant figures. :)
 
  • #3
21,279
4,728
Worked out that the answer for this question is just = 57241.38 J rounded to 3 significant figures. :)
Do you really feel that that is rounded to 3 significant figures?
 
  • #4
Sorry, that my answer from that I rounded to 3 sf so that wasn't my final answer as it is not 3 sf
 

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