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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)

If 450 mL of ethanol density 0789gmol initially at 60C mix with 450 mL of water density 10 gmol initially at 280C in an insulated beaker what is the final tempe class=
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ardni313

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=

[tex]\tt mass=\rho\times V\\\\mass=0.789\times 45=35.505~g[/tex]

mass water =

[tex]\tt mass=1~g/ml\times 45~ml=45~g[/tex]

then the heat transfer :

[tex]\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[/tex]

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