30ml of pure H2O at 280K is mixed w/ 50mL of pure H2O at 330K.What is the final temp. of the mixture?

Question

please explain process.

Answer 1

The heat lost by the warmer water will equal the heat gained by the cooler water. The change in the heat will equal
q = (s)(m)(T1 - T2) where q is the heat change (units of Joules (J)), s is the specific heat of water (4.184 J/gK), m is the mass of the water in grams, and T1 and T2 are the initial and final temperatures of the water samples in Kelvin. Assume the density of water = 1.00g/mL.

(4.184J/gK)(30g)(Tf - 280K) = (4.184J/gK)(50g)(330K - Tf)
Tf is the final temperature of the combined solutions.

Solving for Tf:
Tf = [50(330) + 30(280)] / 80 = 311K