In the first hour, the balloon will increase in volume by two-fifths of 500cm³ which is (2/5)*500cm³ = <<2/5*500=200>>200cm³

After the first hour, it will have a volume of (500+200)cm³ = <<(500+200)=700>>700cm³

In the second hour, it will increase in volume by two-fifths of 700cm³ which is (2/5)*700cm³ = <<2/5*700=280>>280cm³

After the second hour, it's volume will be (700+280)cm³ = <<(700+280)=980>>980cm³

# Answer

980
<|im_start|>system
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<|im_start|>user
A special balloon increases by two-fifths of its previous volume every hour when placed under water. If its original volume is 500cm³, what will its volume be after 2 hours underwater?
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
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user
980
A special balloon increases by two-fifths of its previous volume every hour when placed under water. If its original volume is 500cm³, what will its volume be after 2 hours underwater?
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