[skeemask]

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?

[skeemask]

Can anyone please help?

[mike]

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 =

[skeemask]

Yes, but all they'd do in the equation is cancel out, still leaving the answer at 8.36. By the way, I don't even know if my setup is correct...

[Borek]

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

[skeemask]

Sry, I didn't catch that. Uhm...can you show me how you would approach it, I'm still a little confused.

[Borek]

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.

[skeemask]

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

[Borek]

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.