Answer:
Explanation:
First of all, I used the specific heat of water as 4182 J/(kgC) and the specific heat of ethyl alcohol (EtOH) as 2440 J/(kgC); that means that we need the masses in kg, not g.
120.g = .1200 kg of ethyl alcohol. Now for the formula:
[tex]t_f=\frac{(m_{H2O}*spheat_{H2O}*temp_{H2O})+(m_{EtOH}*spheat_{EtOH}*temp_{EtOH})}{(m_{H2O}*spheat_{H2O})+(m_{EtOH}*spheat_{EtOH})}[/tex] where spheat is specific heat.
Filling that horrifying-looking formula in with some values:
[tex]16.0=\frac{(x*4182*20.0)+(.1200*2440*10.0)}{(x*4182)+(.1200*2440)}[/tex] and
[tex]16.0=\frac{83640x+2928}{4182x+292.8}[/tex] and
16(4182x + 292.8) = 83640x + 2928 and
66912x + 4684.8 = 83640x + 2928 and
1756.8 = 16728x so
x = .105 kg and the amount of water added is 105 g
