Question

If 45.0 mL of ethanol (density =0.789g/mol) initially at 6.0°C mix with 45.0 mL of water (density =1.0 g/mol) initially at 28.0°C in an insulated beaker, what is the final temperature of the mixture, assuming that no heat is lost? (CetOH =2.42 J/G C)

Answer 1

The final temperature of the mixture : 21.1° C  

Further explanation  

The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released  

Q in(gained) = Q out(lost)  

Heat can be calculated using the formula:  

Q = mc∆T  

Q = heat, J  

m = mass, g  

c = specific heat, joules / g ° C  

∆T = temperature difference, ° C / K  

Q ethanol=Q water

mass ethanol=

$\tt mass=\rho\times V\\\\mass=0.789\times 45=35.505~g$

mass water =

$\tt mass=1~g/ml\times 45~ml=45~g$

then the heat transfer :

$\tt 35.505\times 2.42~J/g^oC\times (t-6)=45\times 4.18~J/g^oC\times (28-t)\\\\85.922t-515.533=5266.8-188.1t\\\\274.022t=5782.33\rightarrow t=21.1^oC$