When an hot pan on the stove cools down, what happens to its energy?
2 answers:
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Answer:
0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.
Explanation:
Volume of the radiator = 3.6 L
Percentage of antifreeze = 60%
Total volume of anti freeze in radiator = 60% of 3.6 L :
Percentage of water= 40%
Given,optimal cooling of the engine is obtained with only 50% antifreeze.
So, now we want to reduce the percentage of antifreeze from 60% to 50 %
Volume of coolant removed = x
Volume of water added = x
Volume of anti freeze removed = 60% of(x) = 0.6x
Volume of antifreeze left in radiator =50% of 3.6 L =
1.8 Liter is the desired volume of the antifreeze.
Total volume - Removed volume = desired volume
2.16 L - 60% of( x) = 1.8 L
0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.