I finally solved it!! After almost a week of frustration, I finally found a solution to exercise number 37!! Damn!! It really is very amazing how problems look very complex at first but once you get into it, it becomes so darn easy.. "You'll know when you know", as my professor always say. So, I don't wanna waste some time and brag about Naruto (its only just about 12 minutes away before the download for episode 13 is finished!!!!!) and proceed directly with the problem.
Here it is: the very simple which took me a freaking 6 days of frustration before finally coming up with a solution:
The radiator in a car is filled with a solution of 60% antifreeze and 40% water. The manufacturer of the antifreeze suggests that, for summer driving, optimal cooling of the engine is obtained with only 50% antifreeze. If the capacity of the radiator is 3.6L, how much coolant should be drained and replaced with water to reduce the antifreeze concentration to the recommended level?
So, what's this problem all about? Well basically, the author of the problem obviously wants us to find out how much of the coolant solution should be drained and replaced with water so that the total solution will contain 50% water and 50% antifreeze. Solutions (or mixtures), as we all know from our Chemistry subject are made by mixing by two substances together. Examples include mixing of salt and water (called brine) or maybe as simple as a mix orange powder and water (which is orange juice).. uhh.. okay, I've gotta watch Naruto first (YEEEESSSSSSS!!!), the download has already finished :D
AMAZING!! I've just finished watching episode 13 and it just gets better and better each day!! Another minute of it and I could have been carried away by doing a jutsu here in front of the computer and shout KAGE NO BUNSHIN JUTSU!! :D Well, off to another long wait again for the next episode *sigh*
So, where were we? Ah yes, that stupid exercise number 37. As far as chemistry goes, whenever we mix two substances, we can express the concentration of one substance in the solution as Percentage of Concentration of the Substance (C) x Volume of the Mixutre (V). Going back to our problem, we have:
Total concentration of the Antifreeze (CA) = Percentage of Concentration of the Antifreeze (pCA) x Volume of the Coolant Solution (VC)
CA = pCA x VC
CA = 0.6 x 3.6 Liters
CA = 2.16 Liters
Concentration of the Water (CW) = Percentage of Concentration of the Antifreeze (pCA) x Volume of the Coolant Solution (VC)
CW = pCW x VC
CW = 0.4 x 3.6 Liters
CW = 1.44 Liters
From that formula, we now know that the coolant solution is made up of 2.16 liters of water and 1.44 liters of antifreeze. So now, we have to find out 1) how many liters of the coolant solution should be drained from the radiator and 2) replace it with an equal amount of liters of water 3) so that the total concentration of the antifreeze drops down to the recommended level (which is 50% of the coolant solution or 1.8 liters of antifreeze). An assumption that was made on the problem was that the total volume capacity of the radiator is 3.6 L. This means that the amount of coolant that we drained from the coolant should be equal to the amount of water that we will be added to the coolant. Hence, we have:
Amount of coolant to be drained = Amount of Water to be added
From that, we now form the following equations:
Concentration of Water(CW) x (Volume of the Coolant(VC) - Amount of coolant to be drained(z) ) + Amount of Water to be added (z) = Desired Concentration of Water (dCW) x Volume of the Coolant(VC)
tCW x (VC - z) + z = dCW x VC
0.4 x (3.6 L - z) + z = 0.5 x 3.6 L
Before solving for z, I should first what this equation means. If you would recall the problem, we are requred to increase the total concentration of the antifreeze by reducing the coolant to a certain amount and adding water of the same amount that we reduced to the coolant.
tCW x (VC - z) + z translates just that. Notice that we take the total concentration of the coolant that has already been drained (tCW x (VC - z)) and then add the same amount of liters of water to the already drained coolant (tCW x (VC - z) + z). Now, the right hand side of the equation suggests that the amount of coolant drained and water added should be able to produce 50% concentration of water out of the coolant. = 0.5 x 3.6 L says just that. But wait? isn't it that what we want is to have a 50% antifreeze and not 50% water? Well, yes, but basically, if we increase the water concentration to 50%, then it is implicit that the other part of the solution (which is the antifreeze) would be 50% too! Now if we will solve for z, we will have:
0.6z = 1.8 L - 1.44 L
z = 0.6 L
Eureka!! This is the answer! Hence, we must drain 0.6 liters from the coolant and add the same amount of water to the coolant in order for the concentration of the antifreeze to be reduced to 50% and for the water to be increased by 10%.
Simple isn't it? So what made me took 6 days just to solve this simple problem? Well, I've done some soul searching on this one and found out that I was overdoing the entire thing by adding the antifreeze as part of the equation (and partially because half of my mind was monitoring how many minutes left before the download for episode 12 is completed :D). Basically, I've been forcing the antifreeze and water to work at the same time and this made it unclear for the other side of the equation what the condition is that is needed to be satisfied when all the while, I only needed to know how much desired concentration of water is needed in order to solve the equation. So what's the moral then? Not everything that you see is the real thing! Some facts are only there just to confuse you :D
So, now that I've finally gotten over that, it's time to move on to the next numbers. I still have 30 more questions to answer before I can move to section 1.7. But before that, I think I'll go to sleep first :) Till then..
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