3. I was driving through the desert and my car broke down as the sun was setting. After a quick inspection, I found...
3. I was driving through the desert and my car broke down as the sun was setting. After a quick inspection, I found that a coolant hose had slipped off of the radiator; I fixed it quickly, but all of my coolant was gone, and I needed something to use as coolant if I were to continue my journey. Luckily, I had a large amount of distilled water with me, and I never go anywhere without a gallon of booze. On this particular trip, I had a gallon of booze which is 95 volume% ethanol (C2HO) in water, and I add that to two gallons of pure water. I combined this with my best intentions and a desperate hope to not die in the middle of the desert and put it into the radiator (NOTE: Hopes and Dreams are normally ignored in technical calculations) I decided to sleep through the night and continue my journey in the morning, and as I drifted off, I had a sudden panic about the low temperatures in the desert and their effect on my newly formed antifreeze mixture. I know that the specific gravity of ethanol is 0.789 and that it freezes at -110 °C, while the water freezes at 0 °C. I also know that the freezing point of a mixture between the two is roughly a linear relationship between the mole fraction of ethanol in the solution and the freezing points of the two components. Calculate the molar fraction of ethanol for the mixture I put into my car's radiator. Using a linear interpolation between pure water's freezing point at 0 °C and pure ethanol's freezing point at -110 °C, as a function of molar fraction, determine the freezing point of the mixture. Will my car's antifreeze destroy my engine if the nighttime low temperature in the desert is -20 "C?
3. I was driving through the desert and my car broke down as the sun was setting. After a quick inspection, I found that a coolant hose had slipped off of the radiator; I fixed it quickly, but all of my coolant was gone, and I needed something to use as coolant if I were to continue my journey. Luckily, I had a large amount of distilled water with me, and I never go anywhere without a gallon of booze. On this particular trip, I had a gallon of booze which is 95 volume% ethanol (C2HO) in water, and I add that to two gallons of pure water. I combined this with my best intentions and a desperate hope to not die in the middle of the desert and put it into the radiator (NOTE: Hopes and Dreams are normally ignored in technical calculations) I decided to sleep through the night and continue my journey in the morning, and as I drifted off, I had a sudden panic about the low temperatures in the desert and their effect on my newly formed antifreeze mixture. I know that the specific gravity of ethanol is 0.789 and that it freezes at -110 °C, while the water freezes at 0 °C. I also know that the freezing point of a mixture between the two is roughly a linear relationship between the mole fraction of ethanol in the solution and the freezing points of the two components. Calculate the molar fraction of ethanol for the mixture I put into my car's radiator. Using a linear interpolation between pure water's freezing point at 0 °C and pure ethanol's freezing point at -110 °C, as a function of molar fraction, determine the freezing point of the mixture. Will my car's antifreeze destroy my engine if the nighttime low temperature in the desert is -20 "C?