Question

If the aluminum block is initially at 25 ∘C, what is the final temperature of the block after the evaporation of the alcohol? Assume that the heat required for the vaporization of the alcohol comes only from the aluminum block and that the alcohol vaporizes at 25 ∘C. The heat of vaporization of the alcohol at 25 ∘C is 45.4 kJ/mol, the specific heat of aluminum is 0.903 J/g⋅∘C

Answer 1

Explanation:

The given data is as follows.

       Heat of vaporization ($\Delta H_{vap}$) = 45.4 kJ/mol

       Specific heat ($C_{p}$) = 0.903 $j/g^{o}C$

Let us assume the alcohol given is $C_{3}H_{8}O$ and its mass is 1.12 g. Also, mass of aluminium block is 73.0 g.

First, calculate the moles of alcohol ($C_{3}H_{8}O$) as follows.

          No. of moles alcohol = $\frac{mass}{\text{molar mass}}$

                                              = $\frac{1.12 g}{60.1 g/mol}$

                                              = 0.0186 mol

So, heat absorbed by the alcohol ($q_{alcohol}$) = heat lost by aluminium ($q_{aluminium}$)

          $n \times \Delta H_{vap}$ = $-m \times C_{p} \times \Delta T$

          $0.0186 mol \times 45.4 kJ/mol = - 73.0 g \times 0.903 J/g^{o}C \times (T_{f} - 25^{o}C)$

              12.71 = $- (T_{f} - 25^{o}C)$

                   $T_{f} = 12.3^{o}C$

Thus, we can conclude that the final temperature of the block is 12.3^{o}C[/tex].

Answer 2

The final temperature of the block after the evaporation of alcohol is  12.3 $\rm ^\circ C$.

The given data has:

The initial temperature of block = $\rm 25\;^\circ C$

Weight of block = 73 g

Weight of alcohol = 1.12 g.

The heat of vaporization of alcohol  ($\Delta$H)= 45.04 kJ/mol

Specific heat of aluminum = 0.903 J/g $\rm ^\circ C$

Since heat energy can not be created nor be destroyed, the heat lost by the aluminum is equal to the heat absorbed by the alcohol.

Heat lost by aluminum = -mc$\Delta$T

Heat absorbed by alcohol = n$\Delta$H

moles of alcohol = $\rm \dfrac{weight}{molecular\;weight}$

moles of alcohol = $\rm \dfrac{1.12}{60.1\;g/mol}$

moles of alcohol = 0.0186 moles

Heat absorbed by alcohol = 0.0186 $\times$ 45.04 kJ/mol

Heat absorbed by alcohol = 0.844 kJ

Heat absorbed by alcohol = Heat lost by aluminum

0.844 kJ =  -mc$\Delta$T

0.844 $\rm \times\;1000$ J = - 73 $\times$ 0.903 J/g $\rm ^\circ C$ $\times$ (final temperature - $\rm 25\;^\circ C$ )

0.844 $\rm \times\;1000$ J = -65.919 (final temperature - $\rm 25\;^\circ C$)

12 = - (final temperature - $\rm 25\;^\circ C$)

Final temperature = 12.3 $\rm ^\circ C$.

The final temperature of the block after the evaporation of alcohol is  12.3 $\rm ^\circ C$.

For more information about the heat of vaporization, refer to the link:

https://brainly.com/question/12625048