How much heat energy is absorbed when 6.83 kg of water is heated from 10.0 0C to 47.0 0C. The specific heat of water is 4.184 J/g 0C.

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asked Apr 24, 2019 184k views

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Robwirving · Answer 1 · 2019-04-28T09:58:38+0000

Heat energy= mass x specific heat x change in temperature
change in temperature=47.0=10.0= 37 degrees
convert mass in G = 6.83x1000=6830g
Heat energy is hence= 6830 x 4.184 j/g/degrees x 37=1057338.64 j or 1057.33864Kj

How much heat energy is absorbed when 6.83 kg of water is heated from 10.0 0C to 47.0 0C. The specific heat of water is 4.184 J/g 0C.

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