30.0 mL of pure water at 282 K is mixed with 50.0 mL of pure water at 343 K. What is the final temperature of the mixture

Assuming density of water = 1 g/mL

mass water = 30.0 mL x 1 g/mL = 30.0 g
mass water = 50.0 mL x 1 g/mL = 50.0 g

specific heat water = 4.18 J/g°C

heat lost + heat gained = 0

30.0 g x 4.18 ( T - 282 ) + 50.0 x 4.18 ( T - 320)=0

solving for T

T = 305.6 K

Ithink it’s 3 as the answer.