Question

If 55.0 mL of ethanol (density=0.789g/mL)) initially at 9.0 ∘C is mixed with 55.0 mL of water (density=1.0g/mL) initially at 28.6 ∘C in an insulated beaker, and assuming that no heat is lost, what is the final temperature of the mixture?

Answer 1

Answer:

Final temperature = 22.21  °C

Explanation:

Heat gain by ethanol = Heat lost by water

Thus,  

$m_{ethanol}\times C_{ethanol}\times (T_f-T_i)=-m_{water}\times C_{water}\times (T_f-T_i)$

Where, negative sign signifies heat loss

Or,  

$m_{ethanol}\times C_{ethanol}\times (T_f-T_i)=m_{water}\times C_{water}\times (T_i-T_f)$

For ethanol:

Density = 0.789 g/mL

Volume = 55.0 mL

Considering the expression for density as:

$Density=\frac {Mass}{Volume}$

So,

So, Mass= Density*Volume = 0.789 g/mL * 55.0 mL = 43.395 g

Initial temperature = 9.0 °C

Specific heat of ethanol = 2.57 J/g°C

For water:

Density = 1.0 g/mL

Volume = 55.0 mL

Considering the expression for density as:

$Density=\frac {Mass}{Volume}$

So,

So, Mass= Density*Volume = 1.0 g/mL * 55.0 mL = 55.0 g

Initial temperature = 28.6 °C

Specific heat of water = 0.450 J/g°C

So,  

$43.395\times 2.57\times (T_f-9.0)=55.0\times 4.186\times (28.6-T_f)$

$111.52515\times T_f-1003.72635=6584.578-230.23\times T_f$

$341.75515\times T_f=7588.30435$

Final temperature = 22.21  °C