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].