Lauren
posted by Chemistry on .
The specific heat of water is at 4.184J/g * degree C. How much heat is evolved, or absorbed, when the temperature of 9.00 mol of liquid water cools from 38 degrees C to 28 degrees C?
How do I start this problem

q = mass x specific heat x delta T.
You will need to change the mols of water to grams. 
wouldnt it be 4.184J/g * degrees C
so 4.184 J/ 162 g H2O * 10 degrees C = ? 
mols = grams x molar mass
9.00 mols H2O x (18.015 g/mol) = ?? g H2O
q = g H2O x 4.184 J/g*C x delta T
and delta T = (Tfinal  Tinitial) = (3828) = 10.
The answer will be a negative number which means that heat is evolved (and not absorbed). 
6783.72 J

ok. Actually, if we round the numbers, the last 2 should be a 3. However, if you are concerned about the number of significant figures, the 9.00 limits them to three; therefore, the 6383.73 J would be written as 6.38 x 10^3 Joules.