[candiishop]

If 60.0g Al metal at 100 degrees celsius were added to 60.0g H20 at 18 degrees celsius in a calorimeter of heat capacity (C) 30.0JK-1, what would the final common temperature be, assuming no heat losses? Use cp for Al = 0.90JK-1g-1 and Cp for H20 = 75.3JK-1mol-1

This is the formula use to answer this question:

-q metal = q H20 + q calorimeter

-c(metal) x m(metal) x delta T = C(H20) x mass H20 x delta T + "cal" x delta T-0.90JK-1g-1 x 60.0g x (Tf- 100)K= 75.3JK-1mol-1 x 60.0g x (Tf-18)K + 30.0JK-1 x (Tf-18)K

-54J(Tf-100) = 4518Jgmol-1 (Tf-18) + 30.0J (Tf-18)

I'm confused with the mol units. Do i get rid of it by multiply by mol/(1.008x2 + 16)gmol-1 ??

Thanks.

[Yggdrasil]

The formula for heat from heat capacity varies depending on what units are given for your C.  If your C is in JK^-1g^-1, then the formula is q = mC?T, where m is the mass of your compound.  However, if you C is in units of JK^-1mol^-1, then the formula is q = nC?T, where n is the number of moles of your compound.