If the aluminum block is initially at 25 ∘C∘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∘C. The heat of vaporization of the alcohol at 25 ∘C∘C is 45.4 kJ/molkJ/mol, the specific heat of aluminum is 0.903 J/g⋅∘CJ/g⋅∘C.

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. Heat of vaporization of the alcohol at 25 ∘C is 45.4 kJ/mol Suppose that 1.12 g of rubbing alcohol (C3H8O) evaporates from a 73.0 g aluminum block.

Explanation:

Heat lost by aluminum block = heat required for vaporization of alcohol

Heat required to vaporize ethanol, H = mass of alcohol * heat of vaporization of alcohol

Mass of alcohol = 1.12 g; molar mass of rubbing alcohol = 60 g/mol

Heat of vaporization = (45.4 kJ/mol)/ 60 g/mol = 0.75666 kJ/g

H = 1.12 g × 0.7566 kJ/g

H = 0.8474 kJ = 847.4 J

Heat lost by aluminum block, Q = -(mass × specific heat capacity × temperature change)

Mass of aluminum block = 73.0 g

Specific heat capacity of aluminum = 0.900 J/g

Temperature change = (Final temperature, T - 25)

Q = -73.0 g × 0.900 J/g × (T - 25)

Q = -65.7 J × (T - 25°C)

Since, Heat lost by aluminum block = heat for vaporization of alcohol

-65.7 J × (T - 25°C) = 847.4 J

T - 25 = 847.4/-65.7

T - 25 = -12.9

T = -12.9 + 25

T = 12.1°C