Heat question

1. Jan 30, 2006

kitkat

I am stuck on this problem...
What would be the final temperature when 100 g of 25 degree C water is mixed with 75 g of 40 degree C water?

I know that mc?T gained = mc? T lost
so (100g) (1.0 cal/g C) (Tf -25) = (75g) (1.0 cal/g C)(40 degrees - Tf)

Where do I go from here?
any push in the right direction would be great.

Kat

2. Jan 30, 2006

vaishakh

100g/75g = 4/3 = (40 - Tf)/(Tf - 25).
If I push you for such a small thing you will fall down dear kid.
Generally take this as a formula when same object of differant amounts at differant temperatures are kept in contact with each other and the temperature of equilibrium temperature is asked - M1/M2 = dT1/dT2.

Anyway do you really need a push to solve the above equation?

3. Jan 30, 2006

kitkat

i guess so

I guess I need to be hit in the head with a hammer cause I just cant "see" it.
My teacher gave me the following formula
(100g) (1.0 cal/g C) (Tf -25) = (75g) (1.0 cal/g C)(40 degrees - Tf)

and I just can't see first what the heck Tf stands for, where the 1.0 cal came from an how to solve it at all. I get to this point and then falter.

4. Jan 30, 2006

Hootenanny

Staff Emeritus
With regards to $T_f$ think about it logically. Go through each parameter in your equation. $T_f$ is the only unresolved variable and therefore must be the thing you are trying to calculate.

As for the 1 cal/g C, this is the specific heat capacity of water. It is usually given in the question or on a data sheet.

5. Jan 31, 2006

kitkat

It was set up wrong!!

I figured it out!
The problem should be set up like so:
(100g) (1.0 cal/g Co ) (Tf- 25 degrees) + (75g) (1.0 cal/g Co) (40 degrees - Tf) = 0
Resulting in this:

100Tf – 2500 + 75Tf – 3000 = 0
175Tf = 2500 + 3000
175Tf = 5500
175Tf/175 = 5500/175
Tf = 31.4

I knew something was missing! Yea!!!! I'm not dumb! Thanks to those who tried to help.
Kat