What is the final temperature of the solution assuming that no heat is lost?

Question

A solution is made by mixing 249.0mL of ethanol initially at 10.7°C with 249.0 ML of water initially at 24.7°C. The density of ethanol is 0.789g/mL and the density of water is 1.00g/mL. The specific heat of ethanol is 2.46J/g°C and the specific heat of water is 4.184J/g°C.

Answer 1

Heat gained by the ethanol MUST EQUAL heat lost by the water (heat always goes from warmer to cooler)

Heat = q = mC∆T

m = mass; C = specific heat; ∆T = change in temperature

heat gained by ethanol (EtOH) = q = mC∆T

q = (249.0 ml x 0.789 g/ml)(2.46 J/gº)(10.7º - Tf)

heat lost by water (H2O) = q = mC∆T

q = (249.0 ml x 1.0 g/ml)(4.184 J/gº)(Tf - 24.7º)

Setting these two equations equal, we have...

(249.0 ml x 0.789 g/ml)(2.46 J/gº)(10.7º - Tf) = (249.0 ml x 1.0 g/ml)(4.184 J/gº)(Tf - 24.7º)

Solving for Tf (final temperature) we have ...

5171 - 483Tf = 1042Tf - 25733

1525Tf = 30904

Tf= 20.3º

(be sure to check the math)