# Calculating Final Temperature

## Homework Statement

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).

## Homework Equations

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

## 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:

## Answers and Replies

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

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.

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

--

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?

mgb_phys
Homework Helper
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)

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

mgb_phys