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## 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.

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