Asked by Chemistry

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
Answered by DrBob222
q = mass x specific heat x delta T.
You will need to change the mols of water to grams.
Answered by Chemistry
wouldnt it be 4.184J/g * degrees C

so 4.184 J/ 162 g H2O * 10 degrees C = ?
Answered by DrBob222
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) = (38-28) = -10.
The answer will be a negative number which means that heat is evolved (and not absorbed).
Answered by Chemistry
-6783.72 J
Answered by DrBob222
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.
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