f the ΔH of fusion for aluminum is 10.9 kJ/mol and the ΔH of vaporization is 284 kJ/mol, then what is the change in entropy upon melting down an aluminum can that weighs 18.9 grams?
To determine the change in entropy upon melting down an aluminum can, we'll need to use the following equation:
ΔS = ΔH / T
where:
ΔS is the change in entropy (in J/mol·K),
ΔH is the enthalpy change (in J/mol),
T is the temperature (in K).
First, we need to convert the mass of the aluminum can from grams to moles. To do this, we’ll use the molar mass of aluminum (26.98 g/mol).
Moles of aluminum can = Mass of aluminum can / molar mass of aluminum = 18.9 g / 26.98 g/mol
Now, let's calculate the change in entropy for each step (fusion and vaporization) separately:
For fusion:
ΔH_fusion = 10.9 kJ/mol = 10.9 × 10^3 J/mol
For vaporization:
ΔH_vaporization = 284 kJ/mol = 284 × 10^3 J/mol
Now, we can calculate the change in entropy for fusion (ΔS_fusion) and vaporization (ΔS_vaporization) using the given equations:
ΔS_fusion = ΔH_fusion / T
ΔS_vaporization = ΔH_vaporization / T
Next, we need to calculate the final change in entropy (ΔS_total) by adding the individual changes in entropy:
ΔS_total = ΔS_fusion + ΔS_vaporization
Finally, we can substitute the known values into the equations to find the change in entropy.
Make sure to convert the temperature to Kelvin (K) by adding 273.15 to the temperature in Celsius.
Note: I apologize, but you haven't provided the temperature at which the aluminum can is being melted. Please provide that information so we can calculate the change in entropy accurately.