word problem-show your math

1) A 3.4 liter '97 4runner with no rear heater has coolant which tests good only to 5 degrees F. In order to make it test to 34 below zero, what volume of diluted coolant must be drained to add the equivalent amount of undiluted antifreeze?

2) A train leaves Kansas City at 3:10 pm heading west going 62 miles per hour, and another train leaves Denver at 3:23 pm heading east at 72 mph...
 
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a good follow up question there is how old is your coolant? It might be worth it to just change it all out so you've got not only the right concentration, but brand new fluid.
 
I would do drain and fills and monitor after each one which includes a drive to be sure the stat opens. Going to take at least two, maybe three, depending on how much drains from rad each time. I use Toyota pink coolant.
 
I'm not changing my coolant until spring, at which time I will be doing a thorough flush and switching to pink. It was partially changed when I did my timing belt, which resulted in the current word problem.

That's 2 F's so far...:cowboy:
 
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5*F protection means your coolant is 'acting' like 35-40% coolant to 60-65% water. You at least 50/50 up to 60/40 the other way if you really think you are going to deal with those low temps. I'd probably go drain a full gallon and add a full gallon of full-strength. You may even want to get 55% pre-mix Pink when you swap next year if you can get it from Canada.

I deal with high summer temps more often than freezing so I always run under 50% coolant.

-Charlie
 
Now we can do some math.

motorweek-l1-chart.jpg


So I have 9 quarts of coolant that is good to 5F. This gives is a mixture of 30% antifreeze to 70% water, according to the graph. So I have 9 qts x .3 = 2.7 quarts of antifreeze in the vehicle now.

I need 9 quarts good to lets say 40 below, the graph tells us we need a solution that is about 52% antifreeze and 48% water. So I need 9 qts x .52 = 4.68, lets say 4.7 quarts. I need to get 2 more quarts of antifreeze in there.

But if I drain 2 quarts I will be removing .6 quarts of antifreeze, which means I would need to add 2. 6 quarts antifreeze instead of 2, and there won't be enough room. So I will drain 3 quarts, add 3 quarts, and call it good enough because I don't want to keep doing ever smaller calculations. :cowboy:
 
This is a trick question because undiluted antifreeze has a higher freezing point than 50/50 dilution.

You could have a high freezing point either because you have too much antifreeze and not enough water, or too much water and not enough antifreeze.

So there is no way to answer this question without knowing the ratio of the coolant in the radiator, because that freezing point of 5F could be achieved with 2 different ratios that would require different actions
 
Did you know coolant can go acidic? This is one reason why we have the change intervals. The other reason is contamination. Slowly the coolant can get contaminated and that's is the 2nd reason. When it goes acidic it eats components and that's another why the head gaskets go and the water pump fails. You see a lot more oxidation on components with acidic coolant. This is why its best to use only pink and red Asian coolant as its designed for the engine coolant system. :-)
 
This is a trick question because undiluted antifreeze has a higher freezing point than 50/50 dilution.

You could have a high freezing point either because you have too much antifreeze and not enough water, or too much water and not enough antifreeze.

So there is no way to answer this question without knowing the ratio of the coolant in the radiator, because that freezing point of 5F could be achieved with 2 different ratios that would require different actions
This is just one part of why I said there is incomplete information and thus straight algebra doesn't work. There's no reason to do just 3 quarts when you can do a full gallon and not have a random quart hanging around in a bottle just to get thrown away later...

-Charlie
 
This is a trick question because undiluted antifreeze has a higher freezing point than 50/50 dilution.

You could have a high freezing point either because you have too much antifreeze and not enough water, or too much water and not enough antifreeze.

So there is no way to answer this question without knowing the ratio of the coolant in the radiator, because that freezing point of 5F could be achieved with 2 different ratios that would require different actions

Good point, but you could have asked. It was definitely not too much antifreeze, it was whatever was left after I drained the radiator but not the block, plus distilled water. Thanks for playing!
 
Did you know coolant can go acidic? This is one reason why we have the change intervals. The other reason is contamination. Slowly the coolant can get contaminated and that's is the 2nd reason. When it goes acidic it eats components and that's another why the head gaskets go and the water pump fails. You see a lot more oxidation on components with acidic coolant. This is why its best to use only pink and red Asian coolant as its designed for the engine coolant system. :-)

And that's why I planned to switch to pink but too many other jobs piled up on me so its not getting done until spring now. And since I will have no use for that extra quart of green come spring I will probably just add the whole gallon. I already made the switch to pink on my other Toyota.
 
1) A 3.4 liter '97 4runner with no rear heater has coolant which tests good only to 5 degrees F. In order to make it test to 34 below zero, what volume of diluted coolant must be drained to add the equivalent amount of undiluted antifreeze?

2) A train leaves Kansas City at 3:10 pm heading west going 62 miles per hour, and another train leaves Denver at 3:23 pm heading east at 72 mph...

1) You are at a 30% dilution. You want to be at 50%, both per this chart : https://www.fcsdchemicalsandlubricants.com/quickref/ethylene.pdf

So, you want to achieve .3x + y = 1.05 gallons

and

x + y = 2.1 gallons.

where x is the remaining 30% coolant and y is the undiluted coolant you add.

multiply top equation by 2: .6x + 2y = 2.1 gallons.

Thus: .6x + 2y = x + y (= 2.1 gallons)

Subtract y from both sides

.6x + y = x

subtract .6x from both sides

y = .4x

so: .4x + x = 2.1 gallons = 1.4 x

x = 2.1/1.4 = 1.5 gallons of 30%

so you need to take out .6 gallons, and you add .6 gallons of 100% and you are left with 1.5 gallons of 30% containing .45 gal of 100%, for a total concentrate of (.6 + .45) gal or 1.05 gal of your capacity of 2.1 gallons.

I could do #2 but I am not really interested.

EDIT: For fun, we...well, my computer jock son asked ChatGP to calculate the answer. She blew it. Her answer of .42 neglected that you throw away 30% concentrate with the old coolant removed. Impressively, she admitted I was right and proceeded to get the same answer I did. She also did not know the coolant capacity of a '97 3.4L with no rear heater so I had to give her that.

We turned her loose on #2. She made short work of it. The trains will crash head on (or pass each other if they happen to be on parallel tracks) at 5:45 pm 155 miles east of Denver.
 
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I knew the math guy would show up eventually lol. Good work! I already added 1 gallon, but now I have a formula if there's ever a next time.

I hope nobody is on that train...:)

1) You are at a 30% dilution. You want to be at 50%, both per this chart : https://www.fcsdchemicalsandlubricants.com/quickref/ethylene.pdf

So, you want to achieve .3x + y = 1.05 gallons

and

x + y = 2.1 gallons.

where x is the remaining 30% coolant and y is the undiluted coolant you add.

multiply top equation by 2: .6x + 2y = 2.1 gallons.

Thus: .6x + 2y = x + y (= 2.1 gallons)

Subtract y from both sides

.6x + y = x

subtract .6x from both sides

y = .4x

so: .4x + x = 2.1 gallons = 1.4 x

x = 2.1/1.4 = 1.5 gallons of 30%

so you need to take out .6 gallons, and you add .6 gallons of 100% and you are left with 1.5 gallons of 30% containing .45 gal of 100%, for a total concentrate of (.6 + .45) gal or 1.05 gal of your capacity of 2.1 gallons.

I could do #2 but I am not really interested.

EDIT: For fun, we...well, my computer jock son asked ChatGP to calculate the answer. She blew it. Her answer of .42 neglected that you throw away 30% concentrate with the old coolant removed. Impressively, she admitted I was right and proceeded to get the same answer I did. She also did not know the coolant capacity of a '97 3.4L with no rear heater so I had to give her that.

We turned her loose on #2. She made short work of it. The trains will crash head on (or pass each other if they happen to be on parallel tracks) at 5:45 pm 155 miles east of Denver.
 

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