Please help me out on this. I've been at it for an hour.

A 150.0g sample of a metal at 75.0^oC is added to 150.0 g of H₂O at 15.0^oC. The temperature of the water rises to 18.3^oC. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water.

The answer is supposed to be s = 0.25 J/g x ^oC

What I did was I let delta H be equal and wrote:
(s)(150.0 g )(3.3^oC)=(4.18)(300.0 g)(3.30^oC)

with s = 8.36 which is not even close. Can anyone please tell me what I did wrong?

Ok, so firstly the final temperature of both metal and water should be the same in theory, so the dT is not 3.3 for both the metal and the water:

75.0 - 18.3 =

Yes, but all they'd do in the equation is cancel out, still leaving the answer at 8.36.

No. Heat lost = heat gained. You have two different ?T values here, just like mike wrote - and as they are different, they don't cancel out

Final temperature is 18.3, so water gained enough heat to be warmed by 18.3-15.0 deg, while metal lost enough heat to be cooled by 75.0-18.3 deg. Calculate amount of heat involved in both cases - and compare them as they have to be equal.

What about the masses, is my equation okay in terms of the masses? 150 or 300?

No. 300 g is total mass of the system. To calulate amounts of the heat lost or gained by the individual system components you have to use their individual masses.