Thermal Equilibrium of Hot Iron & Water

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Homework Help Overview

The problem involves determining the common temperature of a hot iron nail and water when they reach thermal equilibrium. The subject area includes thermodynamics and heat transfer principles.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the principle of conservation of energy by equating the heat lost by the iron nail to the heat gained by the water. Some participants question the setup of the temperature equation, particularly the treatment of the initial temperature of the water.

Discussion Status

Participants are actively engaging with the problem, with some providing feedback on the original poster's approach. There is a suggestion to reconsider the temperature difference in the calculations, indicating a productive direction in the discussion.

Contextual Notes

There is mention of the heat capacities of both the iron and water, as well as the initial temperatures, which are critical for the calculations. The original poster expresses confusion regarding a negative temperature result, highlighting a potential misunderstanding in the application of the equations.

TT0
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Homework Statement


A hot (800 K) iron nail of mass 150 g is placed into a 1.0 kg of water at 25°C. What is the common temperature of the 2 objects once they have reached thermal equilibrium? Assume no change of state of the water.

Heat capacity of iron nail is 450 J/kg/K
Heat capacity of water is 4138 J/kg/K

Homework Equations


Q=mc∆T

The Attempt at a Solution


Q lost in nail=Q gained in water

let T = final temperature

0.15*450*(800-T)=1*4138*(T+25+273)
T=-280.38 K

I don't know why it is negative. Can someone explain what I did wrong?
 
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(T+25+273) should be a difference between the final temperature and initial temp.
 
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I see thanks!
 
Is the answer 33.06 C?
 
Seems reasonable
 
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Thanks again!
 

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