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) = (38-28) = -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.

To solve this problem, you'll need to use the equation for heat transfer:

Q = m * c * ΔT

Where:
Q is the heat transferred (in Joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g·°C)
ΔT is the change in temperature (in °C)

To start the problem, we need to determine the mass of water in grams. This can be done using the molar mass of water.

The molar mass of water (H2O) is approximately 18.015 g/mol. Therefore, the mass of 9.00 mol of water can be calculated as:

9.00 mol * 18.015 g/mol = 162.135 g

Now that we have the mass (m) and the specific heat capacity (c), we can calculate the heat transfer (Q). The change in temperature (ΔT) is given as a decrease from 38°C to 28°C, which is -10°C:

Q = 162.135 g * 4.184 J/g·°C * (-10°C)
Q = -6806.1056 J

The negative sign indicates that heat is being lost to the surroundings, as the water is cooling down. Therefore, the heat evolved when the temperature of 9.00 mol of liquid water cools from 38°C to 28°C is approximately -6806.1056 J.