Chemistry help.....

Question

A solution is made by mixing  238.0 mL of ethanol initially at 11.2 ∘C with 238.0 mL of water initially at 24.2 ∘C. What is the final temperature of the solution assuming that no heat is lost? The density of ethanol is 0.789 g/mL and the density of water is 1.00 g/mL. The specific heat of ethanol is 2.46 J/g·°C and the specific heat of water is 4.184 J/g·°C.

Answer 1

For this problem, we will use q = mC∆T where

q = heat

m = mass

C = specific heat

∆T = change in temperature

We also will use the concept of conservation of energy so that the heat lost by the water at 24.2º will be equal to the heat gained by the ethanol at 11.2º (heat goes from hot object to cold object).

Heat lost by water = q = mC∆T

m = mass of water = 238.0 ml x 1.00 g/ml = 238 g

C = 4.184 J/gº

∆T = 24.2º - Tf where Tf is the final temperature

q = (238 g)(4.184 J/gº)(24.2 - Tf)

Heat gained by ethanol = q = mC∆T

m = mass = 238.0 ml x 0.789 g/ml = 188 g

C = 2.46 J/gº

∆T = Tf - 11.2º

q = (188 g)(2.46 J/gº)(Tf - 11.2)

Setting the two equations equal to each other (conservation of energy):

(238 g)(4.184 J/gº)(24.2 - Tf) = (188 g)(2.46 J/gº)(Tf - 11.2)

24098 - 996Tf = 462Tf - 5174

1458Tf = 29272

Tf = 20.1ºC as the final temperature of the mixture