Specific Heat Problem (did something wrong?)

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The discussion focuses on a specific heat problem involving a heated iron mass added to water, with the goal of calculating the initial temperature of the iron. The user initially set up the equation incorrectly, leading to an unrealistic negative temperature result. Another participant pointed out that the negative sign was missing in the heat transfer equation, which is crucial for correctly balancing the heat lost by water with the heat gained by iron. After acknowledging the mistake, the user expressed gratitude for the clarification. This highlights the importance of correctly applying the principles of heat transfer in thermodynamic calculations.
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!
 
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It looks like you lost the negative sign. It should be
$$55 \cdot 0.451\cdot(42.7-X) = -100\cdot 4.184\cdot (42.7-20)$$
 
NFuller said:
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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