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.

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MRROYALGO MRROYALGO · Answer 1

The temperature of an object is how cold or hot it is.

The final temperature is 12.3 degrees Celsius

The given parameters are:

[tex]\Delta H_{vap} = 45.4\ kJ/mol[/tex] --- the heat of vaporization

[tex]C_p = 0.903J/g^oC[/tex] --- the specific heat of Aluminum

[tex]m = 1.12g[/tex] --- the mass of alcohol

[tex]M = 60.1g/mol[/tex] -- the molecular mass of alcohol

Start by calculating the moles of alcohol using:

[tex]Moles = \frac{Mass}{Molecular\ Weight}[/tex]

So, we have:

[tex]n = \frac{1.12g}{60.1g/mol}[/tex]

[tex]n= 0.0186\ mol[/tex]

Next, calculate the amount of heat absorbed by the alcohol using:

[tex]n * \Delta H_{vap} = -m_{alcohol}* C_p * \Delta T[/tex]

The mass of alcohol is 73 grams. So, we have:

[tex]0.0186 * 45.4 = -73* 0.903 * \Delta T[/tex]

Evaluate the products

[tex]0.84 = -65.92* \Delta T[/tex]

Divide both sides by -65.92

[tex]\Delta T = -0.0127[/tex]

Express as degrees Celsius

[tex]\Delta T = -12.7[/tex]

Also, we have:

[tex]\Delta T = T_i - T_f[/tex] --- the difference in the initial and the final temperatures.

This gives

[tex]-12.7 = T_f - 25[/tex]

Collect like terms

[tex]T_f = 25 - 12.7[/tex]

[tex]T_f = 12.3[/tex]

Hence, the final temperature is 12.3 degrees Celsius