A 42.5 g piece of aluminum (which has a molar heat capacity of 24.03 j/ocmol) is heated to 82.4oc and dropped into a calorimeter containing water (specific heat capacity of water is 4.18 j/goc) initially at 22.3oc. the final temperature of the water is 24.9oc. calculate the mass of water in the calorimeter. ignore significant figures for this problem.

1. Use the equation q = nC∆T

n = mols of aluminum = 42.5g / 6.98g/mol

C = molar heat capacity = 24.03 j/Cmol

∆T = change in T from what it was before placed in calorimeter to after =

24.9C-82.4C (because water final and metal will be same temp.)

plug in to calculate q of metal =

q = (42.5/6.98) * (24.03J) (24.9-82.4)

q = - 2176.5498 J

qmetal = - q of water

plug in values for water

- (-2176.5498 J) = mC∆T (m = mass of water in grams)

m = q/C∆T

∆T=24.9-22.3 C = 2.6

2176.5498 J / (2.6x4.184) = m = 200.2714 g