The radiator in a car is filled with a solution of 75 per cent antifreeze and 25 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 3.5 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent?
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To find out how much coolant needs to be drained and replaced with pure water, follow these steps:
Step 1: Determine the current amount of antifreeze in the radiator.
The radiator contains a solution of 75% antifreeze and 25% water. If the capacity of the radiator is 3.5 liters, then the current amount of antifreeze in the radiator can be calculated as:
Current amount of antifreeze = 75% of 3.5 liters
Step 2: Determine the desired amount of antifreeze in the radiator.
For optimal cooling, the manufacturer suggests a 50% antifreeze concentration. To find the desired amount of antifreeze in the radiator, calculate:
Desired amount of antifreeze = 50% of 3.5 liters
Step 3: Calculate the difference between the current and desired amounts of antifreeze.
Difference = Current amount of antifreeze - Desired amount of antifreeze
Step 4: Determine the volume of coolant that needs to be replaced with pure water.
Since the coolant mixture is being adjusted by draining and replacing it with pure water, the volume of coolant that needs to be removed can be determined by calculating:
Volume of coolant to be replaced = Difference / 0.75
Step 5: Calculate the volume of pure water that needs to be added.
Since the volume of coolant to be replaced is equal to the volume of pure water added, the volume of pure water needed can be calculated as:
Volume of pure water = Volume of coolant to be replaced
By following these steps, you can find the amount of coolant (in liters) that needs to be drained and replaced with pure water to reduce the antifreeze concentration to 50%.