Strong sunshine bombards the earth with about 1 kJ.m-2.s-1
Calculate the maximum mass of
pure ethanol that can be vaporized in 10. min from a beaker left in strong sunshine, assuming
the surface area of the ethanol to be 50.cm
2
.
To calculate the maximum mass of pure ethanol that can be vaporized in 10 minutes, we need to find the amount of energy required to vaporize the ethanol and then use the given information about the sunlight energy to determine the mass.
First, we need to find the energy required to vaporize the ethanol. The energy required to vaporize a substance is given by its heat of vaporization (ΔHvap). For ethanol, the heat of vaporization is approximately 38.56 kJ/mol.
Next, we need to calculate the number of moles of ethanol that can be vaporized using the energy provided by the sunlight. We know that the sunlight provides 1 kJ.m-2.s-1, which means it provides 1 kJ of energy per square meter per second.
The area of the ethanol surface is given as 50 cm^2, which is equal to 0.005 m^2. Thus, the energy received by the ethanol surface in 10 minutes (or 600 seconds) is:
Energy = 1 kJ.m-2.s-1 × 0.005 m^2 × 600 s = 3 kJ
Now, we can calculate the number of moles of ethanol that can be vaporized using this energy. To do this, we divide the energy by the heat of vaporization:
Number of moles of ethanol = Energy / ΔHvap = 3 kJ / 38.56 kJ/mol
Finally, we can calculate the maximum mass of ethanol that can be vaporized using the molar mass of ethanol, which is approximately 46.07 g/mol:
Mass of ethanol = Number of moles × molar mass = (3 kJ / 38.56 kJ/mol) × 46.07 g/mol
Calculating this, we get the maximum mass of pure ethanol that can be vaporized in 10 minutes from the beaker left in strong sunshine.