30.0 mL of pure water at 282 K is mixed with 50.0 mL of pure water at 328 K. What is the final temperature of the mixture

Explanation:

As the 50mL of water are in a big temperature will give heat to the other sample of water.

With the equation:

Q = m*C*T

Where Q is heat, m  is mass of the sample, C is specific heat of the sample (4.184J/molK for pure water) and T is temperature in K

We can determine the energy of each sample of water:

Sample 1:

Q = 30.0g*4.184J/molK*282K

Mass is 30.0g because density of water is 1g/mL

Q = 35396.6J

Sample 2:

Q = 50.0g*4.184J/molK*328K

Q = 68617.6J

Total energy is:

Q = 104014.2J

As total mass is 30+50 = 80g, final temperature is:

104014.2J = 80.0g*4.184J*T

310.7K is final temperature of the mixture

yes it is balanced