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Energy Assignment: need a step in the right direction

  • Thread starter Phan
  • Start date
33
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1. Homework Statement
A 5.0g piece of iron at 75.0C is added to 150.0g of water at 15.0C. What is the final temperature of the iron and the water? (use c=0.444J/g*C for iron).


2. Homework Equations
Q = mc(T2-T1)


3. The Attempt at a Solution
In all honesty, I have no idea how to start this question at all. Most of the other questions on this assignment deal with enthalpy changes, but this one seems to be missing some information that is needed to solve it. Previous questions generally gave another variable, like the Q (heat change) value for either one of the iron or the water. Yet, I am missing more than 2/3 of the variables in the Q=mcdeltaT formula, so I have no idea what I can calculate.

If anyone can point me in the right direction (for a hopefully straightfoward problem), it would be much appreciated. Thanks.
 

Answers and Replies

Mapes
Science Advisor
Homework Helper
Gold Member
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It appears that the iron's going to cool down and the water's going to heat up. If energy is conserved, what can you say about the thermal energy lost by the iron vs. the thermal energy gained by the water?
 
33
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It appears that the iron's going to cool down and the water's going to heat up. If energy is conserved, what can you say about the thermal energy lost by the iron vs. the thermal energy gained by the water?
So...

Qiron = -Qwater or vice versa?
Do I set them equal to each other and then solve?
 
138
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My brother asked me the same question about a month ago, LOL.

Yes, Qiron + Qwater = 0

Solve for T2 (the final temperature).
 
33
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My brother asked me the same question about a month ago, LOL.

Yes, Qiron + Qwater = 0

Solve for T2 (the final temperature).
I am assuming that I am doing this because the total heat change between the two is 0?
Well, here is what I did (I suck as using symbols right now, so bear with me for the delta Ts):


Qiron + Qwater = 0
mc(T-T1) + mc(T-T1) = 0

[5.0(0.444)(T-75)] + [150(4.184)(T-15)] = 0
[2.22T-166.5] + [627.6T-9414] = 0
[629.82T - 9580.5] = 0
T = 9580.5/629.82
T = 15.21C

Yet, this answer seems very small... so did I mess up in my algebra somewhere?
 
33
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I just consulted a friend, and he said to use:

Qiron=Qwater...

I'm confused now, as this nets me less that 15C?
 
Last edited:
138
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Yes, T = 15.2 C. Think about it carefully: you have 150 g of water and put only 5 g of iron at 75.0 C, so the temperature isn't going to increase by much.

Energy is conserved: the heat lost by the iron is transferred to the water, therefore Qwater = -Qiron, i.e., Qwater + Qiron = 0.
 

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