- #1

kirsten_2009

- 137

- 2

## Homework Statement

A 1.22 kg piece of iron at 126.5 degrees Celsius is dropped into 981 g of water at 22.1 degrees Celsius. The temperature rises to 34.4 degrees Celsius. What will be the final temperature if this same piece of iron at 99.8 degrees celsius is dropped into 350 mL of glycerol, CH2OHCHOHCH2OH, at 26.8 degrees Celsius? For glycerol, density = 1.26 g/mL, C_p = 219 J/K x mol.

## Homework Equations

q = m x C x (Tf-Ti)

## The Attempt at a Solution

So, I know that the specific heat of water is 4.18 J/g x C, so I could find "q" for water (50437.134) and since the heat gained by the surroundings is equal to what is lost by the system I could also say that "q" for water is the same as "q" for iron but negative (-50437.134). Now, I could calculate the specific heat of iron (0.449 J/g x C). But now I'm stuck because I don't have "q" for the iron that is then dumped into glycerol and I don't have Tf so how could I calculate it? Or, is the value of "q" for iron the same throughout the process? Help is much appreciated! :)