Question

Suppose that 1.06 g of rubbing alcohol (C3H8O) evaporates from a 72.0 g aluminum block. 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.

Answer 1

Answer:

The final temeprature of the block is 12.71 °C

Explanation:

Step 1: Data given

Mass of C3H8O = 1.06 grams

Mass of aluminium block = 72.0 °C

The initial temperature of the aluminium block = 25.0 °C

The alcohol vaporizes at 25 ∘C.

The heat of vaporization of the alcohol at 25 ∘C = 45.4 kJ/mol,

The specific heat of aluminum is 0.903 J/g∘C

Step 2: Calculate moles C3H8O

Moles C3H8O = mass C3H8O / molar mass C3H8O

Moles C3H8O = 1.06 grams / 60.1 g/mol

Moles C3H8O = 0.0176 moles

Step 3: Calculate energy

Energy = 45.4 kJ/mol * 0.0176 moles

Energy = 0.799 kJ = 799 J  gained by the alcohol and lost by the Al

Step 4: Calculate the change of temeprature

Q = m*c*ΔT

ΔT = Q / (m*c)

ΔT = 799 J / (72.0*0.903)

ΔT = 12.29 °C

. Step 5: Calculate the final temperature

T2 = 25.0 °C - 12.29 = 12.71 °C

The final temeprature of the block is 12.71 °C