Question

If 55.0 mL of ethanol ) initially at 8.0C is mixed with 55.0 ml of water (density = 1.0g/mL) initially at 28.6C in an insulated beaker, and assuming that no heat is lost, what is the final temperature of the mixture?

Answer 1

Mass of ethanol, m,E = density * volume

                                 = 0.789 g/mL * 55.0 mL

                                 = 43.4 g

Mass of water, m,H2O = 55.0 mL * 1.0 g/mL

                                     = 55.0 g

Given data:

Initial temperature of ethanol is T,E = 8.0 C

Initial temperature of water is T,H2O = 28.6 C

Let, final temperature of mixture be T.

Heat capacity of water is, C,H2O = 4.18 J/gC

Let, heat capacity of ethanol is C,E

As no heat is lost to surroundings, heat gained by ethanol is equal to heat lost by water.

                                   q,E = - q,H2O

                m * C * (T-T,E) = -m * C * (T - T,H2O)

             43.4 g * C * (T - 8.0) = -55.0 g* 4.18 J/gC * (T - 28.6) C

                  43.4 C (T-8.0) = 230 (T-28.6)

Heat capacity of ethanol is not mentioned in the problem, if we know that we can calculate the final temperature.