Another problem dealing with heat transfer

[Rhine720]

Homework Statement

How much Fe at 200.0C must be placed in 100.00g of water at 20C so that the temperature of both will be at 35.0?

Homework Equations

Q=MCdeltaT

The Attempt at a Solution

Energy required to raise water is 6300 J
So I did by:
(keep in mind I used sig figs)
Q=100.00(4.18)15
Q=6300

And then..

I assumed I must plug in 6300 as a negative to the FE to get the amount needed
I did
-6300=x(.449)165
x=-6300/74.1
x=85g

I go to check my work by doing taking out the final temp and trying to solve for it by using the grams

I did so by:
(85).449(Tf-200.0)=(100.0)4.18(35.0-Tf)
38Tf-7600=14600-418Tf
456Tf=22200
And got 48.7 Which is not so close to 35C


Am I checking this wrong or am I doing the problem wrong?

 

[Delphi51]

I think you have done the problem correctly and the problem with the check is that the "35-Tf" should be "Tf - 20".

You know, your work would be so much clearer if you put in a few words like this:
Heat lost by iron = heat gained by water
mC(delta T) = mC(delta T)
m(.449)(165) = 100*4.18*15
m = 100*4.18*15/(.449*165) = 84.6 g

 

[tomcue]

It seems like you have the right approach to solving this problem. However, it is important to note that the specific heat capacity of iron (Fe) is 0.449 J/g°C, not 0.449 J/g. This small mistake could have affected your final answer. Additionally, when plugging in the values for the final temperature, make sure to use the same units (in this case, either °C or K). Using different units can also lead to an incorrect answer. It would be helpful to double check your calculations and units to ensure accuracy.