Question

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.

Answer 1

Answer:

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