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anonymous

  • 4 years ago

Water (2690g ) is heated until it just begins to boil. If the water absorbs 5.31×10^5J of heat in the process, what was the initial temperature of the water?

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  1. anonymous
    • 4 years ago
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    alright lets try solving for delta t really quick \[\frac{5.31x10^5}{(2690g)(4.184)}=\Delta T\] let me know what you get.

  2. anonymous
    • 4 years ago
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    Did you figure out this one?

  3. anonymous
    • 4 years ago
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    No i cant im super confused:/

  4. anonymous
    • 4 years ago
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    you can't divide 5.31x10^5 by 4.184 and then again by 2690?

  5. anonymous
    • 4 years ago
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    that will give you the amount that the water changed, but then you have to subtract that from the final temp of the water to get where you started from

  6. Xishem
    • 4 years ago
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    What equation are you starting with zbay? I think it's helpful to provide that.

  7. anonymous
    • 4 years ago
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    I was using q=C *M Delta T but the euation thing isn't loading up

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