[johnstoian]

"A 25.0 g sample of pure iron at 85°C is dropped into a 75 g sample of water at 25°C. What will be the final temperature of the water-iron mixture?"

Obviously the water will gain energy and the iron will lose energy. If Q=smΔT, then the energy released by the iron has to be equal to the energy gained by the water. Therefor:

(.45J/1g1°C)(25g)(°C)=(4.184J/1g1°C)(75g)(°C)

I'm having problems understanding what goes where the question marks are. Please helps me out :]

[johnstoian]

Crap, I made a typo, the water is 20°C, not 25°C

[ARGOS++]

Dear johnstoian;

Set the final temperature to x °C; that makes ∆T = (T_Inital – x).
Finally solve the equation for x.

I hope to have been of help to you.
Good Luck!
                    ARGOS⁺⁺

[johnstoian]

Which makes it:

(.45J/1g1°C)(25g)(85-x)=(4.184J/1g1°C)(75g)(20-x)

956.25-11.25x=6276-313.8x

302.55x=5319.75

x=17.58

However, that's not the answer I SHOULD be getting, the answer my teacher gave me is 22°C :/

[ARGOS++]

Dear johnstoian;

I got 22.25°C; - But you did a mistake with one sign!

For water you must set ∆T = - (T_Inital – x), because one is spending energy while the other receives energy.

Good Luck!
                    ARGOS⁺⁺

[johnstoian]

AHA! I get what you're saying. Thank you! :]

[ARGOS++]

Dear johnstoian;

You 're welcome!   ─   Soon again.

Good Luck!
                    ARGOS⁺⁺