Suppose that 1.15 g of rubbing alcohol (c3h8o) evaporates from a 65.0-g aluminum block. if the aluminum block is initially at 25 °c, what is the final temperature of the block after the evapo- ration 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 final temperature of the block after the evaporation of the alcohol is T₂ = 10.18°C

The relation used in determining the number of moles of a substance is the substance mass divided by its molar mass.

Mathematically, we have:

Given that:

number of moles = 0.0191 moles

Using the formula for the heat capacity;

Q = mCΔT

(45.4 J/moles × 0.0191 moles) = 65.0 g × 0.900 J/g° C × ΔT

867 J = 58.5 J × ΔT

ΔT = 867 J /58.5 J

ΔT = 14.82

We know that ΔT is the change in temperature;

ΔT = T₁ - T₂

14.82 = 25 - T₂

T₂ = 25 - 14.82

T₂ = 10.18°C

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The final temperature is 10.2 °C

Explanation:

Step 1: Data given

Mass of C3H8O = 1.15 grams

Mass of an aluminium block = 65.0 grams

Initial temperature = 25.0 °C

Molar mass of C3H8O = 60.1 g/mol

Heat of vaporization of the alcohol at 25 °C is 45.4 kJ/mol

Specific heat of aluminium at 25°C = 0.900 J/g°C

Step 2: Calculate moles of C3H8O

Moles C3H8O = mass C3H8O / molar mass C3H8O

Moles C3H8O = 1.15 grams / 60.1 g/mol

Moles C3H8O = 0.0191 moles

Step 3: Calculate heat

Q = 45.4 kJ/mol * 0.0191 moles = 0.867 kJ = 867 Joules

Step 4: Calculate ΔT

Q = m*c*ΔT

⇒ Q = the heat transfer = 867 J

⇒ m = the mass of aluminium = 65.0 grams

⇒ c = the specific heat of aluminium = 0.900 J/g°C

⇒ ΔT = The change of temperature = TO BE DETERMINED

867 J =65.0 g *0.900 J/g°C * ΔT

ΔT = 867 / (0.900*65.0)

ΔT = 14.8

Step 5: Calculate the final temperature

ΔT = T2 - T1

14.8 = 25.0 - T1

T1 = 25.0 - 14.8

T1 = 10.2 °C

The final temperature is 10.2 °C