Question

Determine the final temperature of the liquid when you mix:

m₁ = 1.00 kg water at t₁ = 10.0 °C

and

m₂ = 2.00 kg water at t₂ = 40.0 °C.

Assume that no heat is lost to the environment.

Q = mc $\Delta$ T

Answer 1

Q. Given for water mass mi =1.0kg at Ti = 10.0°C nired with water of mass m₂ = 2.0 kg at T2 = 40.0 let Tf Ps final temperature of mixer. Gair of Heat by Heat loss by Water at Tz Water at T -R CAT, again Q toss h, CAT, where where Heat is 8=MCAT ( = specific Heat of water morte - T.) s-m ( Tp-T.) my Tf -hati Ma Tz - m. Tf mi Tq tm. Te Mit + m₂ T Tf (m, the moti tha Tz mit tm. Ta final temperature of mixer mytM₂ (1.0)10 + (2.0) (40.0) = 30°C 1.07 2.0 The Température of final mixer is Te = 30°c Te 10 + 80 To go go 3 co