If 450 ml of ethanol density 0789 gmoL initially at 70 degr

If 45.0 ml of ethanol density = 0.789 g/moL) initially at 7.0 degre celcius is mixed with 45.0<?xml:namespace prefix = v ns = \"urn:schemas-microsoft-com:vml\" /> <?xml:namespace prefix = o ns = \"urn:schemas-microsoft-com:office:office\" /> of water density = 1.0 g/ml)initially at 28.7 degre celcius in an insulated beaker, and assuming that no heat is lost, what is the final temperature of the mixture

T=

Solution

your solution is this

Heat gained by ethanol = Heat lost by water

+ [m x Cp x (Tfinal - Tinitial)] of ethanol = - [m x Cp x (Tfinal - Tinitial)] of water

mass of ethanol = 45 mL x 0.789 g/mL = 35.505 g
Cp of ethanol at 7 degrees C (check your textbook) = 2.3 J / g.degrees C
Tfinal = final temperature of mixture = required in the problem
Tinitial ethanol = 7 degrees C
mass of water = 45 mL x 1g/mL = 45 grams
Cp of water = 4.184 J/g. degrees C
Tinitial of water = 28.4 degrees C

Substitute all the given data in the equation

+ [35.505 x 2.3 x (Tfinal - 7)] = - [45 x 4.184 x (Tfinal - 28.4)]

Simplify the expression above and solve for Tfinal:

Tfinal = 21.93 degrees C =====> answer