Specific Heat Problem (did something wrong?)

  • Thread starter Lori
  • Start date
  • #1
Lori

Homework Statement


Heated iron with mass of 55.0 g was added to 100 mL of water at 20 degree Celsius. Assuming no energy transfer to the surroundings and that the final temperature of the system is 42.7 degrees C, calculate initial temperature of the iron.

Mass of Iron = 55.0 g
mass of Water = 100g
initial Temp of Water = 20 degrees C
Final temp of Water/Fe = 42.7 degrees
Want: initial temp of Fe

Given Cs of Fe = 0.451 J/gC
Cs of Water = 4.184 J/gC

Homework Equations



heat of Iron = -heat of water (q=-q)

mCT = mCT

The Attempt at a Solution


mCT of Fe = -mCT of water
[/B]
plug in:
X= initial temp of Fe
55 * 0.451*(42.7-X) = 100*4.184*(42.7-20)

When i solve for X , i get a large negative number = -9497.68 which is clearly not the final temperature!

Was wondering where i went wrong!
 

Answers and Replies

  • #2
It looks like you lost the negative sign. It should be
$$55 \cdot 0.451\cdot(42.7-X) = -100\cdot 4.184\cdot (42.7-20)$$
 
  • #3
Lori
It looks like you lost the negative sign. It should be
$$55 \cdot 0.451\cdot(42.7-X) = -100\cdot 4.184\cdot (42.7-20)$$
Wow, didn't realize that. Thanks !
 

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