You have this great idea! You will start a business removing snow from people\'s

Question

You have this great idea! You will start a business removing snow from people's driveways using a propane torch. How much will you have to charge your customers per square foot of driveway in order to make a profit? Will you leave the driveways wet or dry? Why? Please use these estimates and chemical information: Average driveway is 1000 square feet Average snowfall is 6 inches. 12 inches of snowfall equals 1 inch of rainfall. 20 pounds of propane costs $14.00 delta H fusion of water = 6.0 kJ/mol. delta H vaporization of water = 40.7 kJ/mol. Specific heat of water = 4.18 J/g degree C. Heat of combustion of propane: 1 mol provides 2220kJ energy

Explanation / Answer

It asks if we will leave the driveways wet or dry. That seems like it is pretty much up to you, but if you leave them wet, they'll just freeze over again, so I would say that we should probably leave them dry and I will approach the problem accordingly. Let's start by figuring out how much snow/ water we are dealing with. 1000 square ft of driveway. .5 ft of snow = .5 inches of rain = 1/24 ft of rain 1000 ft^2 * 1/24 ft = 1000/24 ft^3 of water (for the average driveway and average snowfall) In cubic meters, this is 1.18 m^3 of water. In liters it is 1180 L. Now we need to figure out how many mols of water this is: Since the density of water is about 1 kg/L, we have 1180 kg (1180000 g) of water in the average driveway. The molar mass of water is 18.0153 g/mol. So the number of moles of water is: 1180000 g / 18.0153 g/mol = 65500 mol Okay, now we can figure out how much energy it takes to melt the snow in the driveway: First we have to convert that snow to water. Then we have to heat that water to 100 degrees C Then we have to convert that water to gas. Each one of these steps requires energy. If we call T the temperature of the snow, it looks like this: (6000 J/mol)*(65500 mol) + (100 C)*(4.18 J / (g*C))*(1180000 g) + (40700 J/mol)*(65500 J/mol) = 3552090000 J = 3552090 kJ How many moles of propane must we burn to get that much energy? 3552090 kJ / 2220 kJ/mol = 1600 mol propane. the molar mass of propane is 44.1 g/mol, so 1600 mol * 44.1 g/mol = 70.56 kg. That is 155.6 pounds of propane. Propane costs $14 for 20 pounds, so 155.6*(14/20) = $108.92 So, just to cover the cost of the combusted propane, you would need to charge $108.92.

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