100.000g of water at 80.0oc was mixed with an equal amount of cool water in a calorimeter. the final temperature of the water in the calorimeter is 50.0oc. what is the heat (q) in kilojoules for the hot water? (specific heat for water is 4.18j/goc).

The answer to your question is: Q = 12540 kJ

Explanation:

Data

mass = 100,000 g

Cp = 4.18 /g°C

T1 = 80°C

T2 = 50°C

Formula

                       Q = (100000)(4.18)( 50 - 80)

                       Q = - 12540000 J

                       Q = 12540 kJ

it's filtration, but this is an assumption so my apologies if i got this