Question

Please help i'm going to throw up from stress

Water at 20.0 °C is mixed with 120.0 g of ethyl alcohol at 10.0 °C in a thermally insulated container. If the final mixture has a temperature of 16.0 °C, how much water was added?

No links istg

Answer 1

Step-by-step 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:

`t_f=((m_(H2O)*spheat_(H2O)*temp_(H2O))+(m_(EtOH)*spheat_(EtOH)*temp_(EtOH)))/((m_(H2O)*spheat_(H2O))+(m_(EtOH)*spheat_(EtOH)))` where spheat is specific heat.

Filling that horrifying-looking formula in with some values:

`16.0=((x*4182*20.0)+(.1200*2440*10.0))/((x*4182)+(.1200*2440))` and

`16.0=(83640x+2928)/(4182x+292.8)` 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