The radiator in a car is filled with a solution of 65 per cent antifreeze and 35 per cent water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50 per cent antifreeze. If the capacity of the raditor is 3.6 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent?
The radiator in a car is filled with a solution of 65 per cent antifreeze and 35 per cent water.
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Let's define some variables 1) V = volume of water the radiator can hold = 3.6L 2) CWinter = concentration of antifreeze for Winter = 65% = .65*3.6 = 2.34L 2a) Concentration of water in Winter = 100-65 = 35% = .35*36 = 1.26L 3) CSummer = concentration of antifreeze for Summer = 50% = .50*3.6 = 1.80L 3a) Concentration of water in Winter = 100-50 = 50% = 1.80L Concentration = Liters of Antifreeze/Volume in radiator For summer we want 1.8L of antifreeze and 1.8L of water. To get 1.8L of antifreeze from the winter mixture we need to drain the radiator to the volume of antifreeze from 2.34L to 1.8L or a ratio of 1.8/2.34 = .769, or a 23.1% reduction. Since you can't separate the water and antifreeze you must reduce the same percent of water, thus you must drain 23.1% of the winter mixture or, 3.6*.231 = .832L. Since you wish the radiator to be full during the summer too, you must add back .832L of water.
Let x be the amount you need to drain and replace with pure water.
65(3.6-x) = 50(3.6)
Why does this set-up work?
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