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
