Question

Question

25.0 mL of ethanol (density = 0.789 g/mL) initially at 7.0 C is mixed with 35.0 mL of water (density = 1.0 g/mL) initially at 25.3 C in an insulated beaker. Assuming that no heat is lost, what is the final temperature of the mixture?

please explain equation step by step I am confused and thank you in advance (my professor isn't helpful )

Answer 1

First we need to calculate how many grams of ethanol and water are present

and,

second we need standard specific heat for ethanol and water for calculation

Mass of ethanol (g) = density of ethanol x volume of ethanol

                                = 0.789 g/mL x 25 mL

                                = 19.725 g

Mass of water (g) = 1.0 g/mL x 35 mL = 35 g

From literature,

specific heat of ethanol = 2.46 J/g.C

and,

specific heat of water = 4.184 J/g.C

Let x be the final temperature of solution then,

Using equation,

q(ethanol) = q(water)

q = heat

or,

mCpdT(ethanol) = mCpdT(water)

m = mass

Cp = specific heat

dT = change in temperature

(19.725 g) (x - 7.0)C (2.46 J/g.C) = (35.0 g) (25.3 - x)C (4.184 J/g.C)

Solving for x,

48.5235 (x - 7.0) = 146.44 (25.3 - x)

x - 7.0 = 3.018 (25.3 - x)

x - 7.0 = 76.3534 - 3.018x

4.018x = 83.3534

x = 20.745 C

thus, the final temperature of the mixture will be 20.745 C.