Respuesta :
Answer : The final temperature of Aluminium is 17° C
Explanation :
The rubbing alcohol evaporates at 25 °C . That means it changes from liquid state to gaseous state. This process is known as Vaporization and the amount of heat it absorbs is known as enthalpy of vaporization.
Step 1 : Find moles of alcohol
We have 1.5 g of C₃H₈O.
Let us find the moles of alcohol.
[tex]Mole = \frac{Mass}{MolarMass}[/tex]
Molar mass of rubbing alcohol is 60.1 g/mol
[tex]mole = \frac{1.5g}{60.1g/mol} = 0.025 mol[/tex]
Step 2 : Find heat absorbed by given amount of alcohol
The enthalpy of vaporization of rubbing alcohol is 45 kJ/mol
The amount of heat absorbed by alcohol can be calculated as follows.
[tex]q_{alcohol} = \bigtriangleup H_{vap} \times moles[/tex]
[tex]q_{alcohol} = \frac{45kJ}{mol} \times 0.025mol = 1.12 kJ[/tex]
So during the process of evaporation, alcohol absorbs 1.12 kJ of heat.
This heat is taken from the aluminium.
Step 3 : Equate the q values for alcohol and aluminium
Therefore we have , [tex]q_{alcohol} = - q_{Al}[/tex]
[tex]q_{Al} = -1.12 kJ[/tex]
Step 4 : Find delta T for Al using the heat value found in the above step
Heat lost by Aluminium can be calculated using following equation
[tex]q = m \times C \times \bigtriangleup T[/tex]
Here q is the amount of heat lost by Al. We need to convert q from kJ to J
[tex]1.12 kJ \times \frac{1000J}{1kJ} = 1120 J[/tex]
C is the specific heat of Al which is 0.9 J/gC
ΔT is the change in temperature of Aluminium.
m is the mass of Al which is 150 g.
Let us plug in the values and solve for ΔT
[tex]-1120J = 150g \times 0.9 J/gC \times \bigtriangleup T[/tex]
[tex]\bigtriangleup T= \frac{-1120J}{0.9J/gC \times 150g}[/tex]
[tex]\bigtriangleup T = -8.3 C[/tex]
Step 5 : Find final temperature using delta T
But ΔT is the difference in final temperature and initial temperature.
[tex]\bigtriangleup T = T_{f} - T_{i}[/tex]
[tex]-8.3 = T_{f} - 25 C[/tex]
[tex]T_{f} = 25 C - 8.3 C[/tex]
[tex]T_{f}= 16.7 C[/tex]
The final temperature of Aluminium is 17° C
