I'm stumped with specific heat can anyone help!?

Question

Im working on a lab and I am stumped on how to find specific heat. I have water and 100g Iron Fe at a Intial tempature of 20c and I place the iron fe in a test tube and heat it inside of a beaker with 150ml of H20 which is at 20c as well. I heat it untill the iron fe and water has reached both 100c and place the Iron Fe into a Calorimeter filled with 100ml of H20 at 20c as well. Once I place the 100g of Iron Fe at 100c into the Calorimeter the tempature of the water goes up to 28c. How am I supposed to find the specific heat of Fe using that senario can anybody please help!!! Thank you in advance

Answer 1

First law of thermodynamics says that you must be able to account for all of the energy in the system. So, your iron lost heat that had to be gained by the water in your calorimeter. So:

q(gained) = -q(lost)

For the left side, the water gained heat so its temperature increased:
q(gained) = (mcDT) for the water
= 100 g (4.184 J/gc)(8c)
That has to equal the heat lost by the iron:
-q lost = -100 (c)( 72)

Setting these two equal and solving for c gives you the specific heat of the iron.

Answer 2

We just did this in my physics class. You need two equations.
First we define Q, which is the total heat transferred in the system of water and iron.

Q=cH2O X Change of Temperature X mas of water, where c is the specific heat of water, or 4.18 J/K-g.

So Q = 4.18 X 150g H20 X 8 = 5, 016 J

Now we change the equation for the iron

c iron = Q/ change of T X mass of iron

c = 5016/ (100-28) x 100g

c= .696

Hope this helps