# Calculating Final Temperature

1. Nov 18, 2009

### TrueStar

1. The problem statement, all variables and given/known data

A piece of iron weighed at 120 grams at 95.0 degrees Celcius is added to 750ml water of 25.0 degrees Celcius. What is the final temperature of the water when it is equilibrated? Assume the dentisy of water is 1 g/ml. (Specific heat of water is 4.184 and specific heat of iron is .449).

2. Relevant equations

q=mc(delta t). In this case the q of iron is equal to the q of water.

3. The attempt at a solution

Here is my setup:

120*.449(x-95)=750*4.184(x-25)

Solving:
53.88x-5118.6=3138x-78450
53.88x=3138x-73331.4
-3084.12x=-73331.4
x=23.85

I know this is wrong, but I'm not sure if it's because I set this up wrong or did my calculations incorrectly.

Last edited: Nov 18, 2009
2. Nov 18, 2009

### mgb_phys

It's a sign error, the change in temperature from 95>x and 25->x must be different signs

3. Nov 18, 2009

### TrueStar

I don't understand what you mean. I know that 23.85 implies that the temperature decreased for both the iron and water, and that is wrong.

4. Nov 19, 2009

### Staff: Mentor

mgb points out that there is a sign error in your original equation.

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5. Nov 19, 2009

### TrueStar

I think I forgot the qin=-qout concept. Let me try again.

120*.449(x-95)=-750*4.184(x-25)
53.88x-5118.6=-3138x+78450
3191.88x=83568.6

x=26.18 degrees Celcius

Is this what you were referring to?

6. Nov 19, 2009

### mgb_phys

Almost,
if one thing is cooling from 95deg to x then the change in T is (95-x)
then if the other is heating form 25 to X the change must be (x-25) which gives the different signs.

Putting a minus on one side is the same thing (or at least has the same effect)

7. Nov 19, 2009

### TrueStar

I thought that in order to do the equation correctly, one must always subtract the initial temperature from the final temperature.

8. Nov 19, 2009

### mgb_phys

The change in temperature defines the heat flow, so heat in and heat out would have different signs.

9. Nov 19, 2009

### TrueStar

I see. Thank you for checking out my question and taking the time to explain it. :)