Thermodynamics: Dropping a hot horseshoe into a pot of water

Jess_18033152

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

2. Homework Equations
Q= mc (change in temp)

3. 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???

Last edited by a moderator:
Related Introductory Physics Homework Help News on Phys.org

Jess_18033152

Worked out that the answer for this question is just = 57241.38 J rounded to 3 significant figures. :)

Chestermiller

Mentor
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?

Jess_18033152

Sorry, that my answer from that I rounded to 3 sf so that wasn't my final answer as it is not 3 sf

"Thermodynamics: Dropping a hot horseshoe into a pot of water"

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving