Quantity of Heat/Specific Heat Water+Iron

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Homework Help Overview

The problem involves determining the final temperature of a system consisting of 100g of water at 20 degrees Celsius and 100g of iron nails at 40 degrees Celsius, utilizing the concept of heat transfer and specific heat capacities.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the heat transfer equations but expresses confusion about the correct setup for the calculations, particularly regarding the initial and final temperatures of the substances involved.
  • Some participants question the assumptions made about the temperatures and the calculations presented, suggesting a need for clarity in defining the final temperature as a common variable.
  • Others suggest re-evaluating the equations used for energy lost and gained, emphasizing the importance of equating these values to find the final temperature.

Discussion Status

Participants are actively engaging with the original poster's attempts, providing feedback and corrections. Some have offered guidance on how to set up the equations correctly, while others have shared their own calculations and results, indicating a collaborative effort to clarify the problem.

Contextual Notes

There are indications of misunderstanding regarding the definitions of initial and final temperatures, as well as the interpretation of energy calculations. The discussion reflects a mix of attempts to solve the problem and clarifications on the setup required for the equations.

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[SOLVED] Quantity of Heat/Specific Heat Water+Iron

Homework Statement


Exact Problem:

What will be the final temperature of 100g of 20 degrees C water when 100g of 40 degrees C iron nails are submerged in it. (The specific heat of iron is .12 cal/g C. Here you should equate the heat gained by the water to the heat lost by the nails.)


Homework Equations



Q = cm(change in)t

Q = quantity of heat

c = specific heat
(water = 4.184J/g or 1 cal/g)
(iron = .502J/g or .12 cal/g)

m = mass
100g + 100g = 200g total substance(water+iron)

t = temperature
(40(final)-20(initial))

The Attempt at a Solution



I feel I slightly understand what I am supposed to do but the part where it says I need to find the heat lost by iron and the heat gained by water I am totally confused about. Heres my attempt: My assumption is that I need to do two different calculations and difference them out. Also if someone could point me in the right direction to be able to post real equations instead of what I have shown that would be so awesome.

Iron First:
Q = cm(Tf-Ti)

Q = .502J*100g(40-20)

Q = 1004J of Energy Lost

Water Second:

q = 4.184*100g(40-20)

Q = 8368J of Energy Gained

Answer:
8368J - 1004J = 7364J of Energy or 1760.04cal/g

I have tried numerous ways to find the temperature difference, lost and gained but I am clueless. Also I can't figure out where to transfer back into celsius. By my calc's it would have gained like 1700 degrees haha. >.> I feel dumb... Anyways TIPS are always appreciated in advance. Constructive critism is appreciated as well, and well anything else that can make me a better poster is welcome too. ^^ thanks!
 
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Your only problem appears to be that you have not read the problem correctly and don't seem to know what is being asked! Your start is correct but why are you using 40 degrees and 20 degrees as "initial" and "final" temperatures for both water and iron?

You also say
"Answer:
8368J - 1004J = 7364J of Energy or 1760.04cal/g "
which you should realize makes no sense. The question was "What will be the final temperature" and that's not even a temperature.

20 degrees is the initial temperature for the water and 40 degrees is the initial temperature for the iron. The final temperature of both will be the same, of course, so just call that "T".

Your equations should be
Iron: Energy lost Q = .502J/g*100g(T- 40) and
Water: Energy gained q = 4.184J/g*100g(T-20) and they must be equal. Solve for T.
 
I see the fault, Thanks a bunch Ivy. You have saved my day and made me feel all the more dumb O.o. That was a joke btw.

Edit: I came out with 32 degrees C as the final for both. Any chance I can get confirmation on this? Ill wait awhile and if nobody posts Ill mark as solved. Thanks again Ivy, your my hero!
 
Last edited:


Oh my GOD

Temperature will be 28 C in equilibrium with water and nail.
If you have eqations ready is simple.
Put them L one = P one and find answer for T.
Nails will lose from 40C - 12C= 28C
Water will gain heat from 20C + 8C = 28C also
very simple and time consuming but some people are fast so it is their own negative to be not able to stop and think for a while to grab the answare in the balls
 

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