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OpenStudy (anonymous):
How much heat is absorbed by 15.5 g of water when its temperature is increased from 20.0°C to 50.0°C? The specific heat of water is 4.184 J/(g°C).
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OpenStudy (aaronq):
q=mC(T2-T1)
OpenStudy (anonymous):
could you explain that further? lol so it would be q=15.5(20-50)?
OpenStudy (aaronq):
close, m=mass, 15.5 g C=specific heat capacity 4.184 J/(g°C) T2= final temp, so 50 T1= initial temp so 20
OpenStudy (anonymous):
okay so then i would do it this way: q=15.5*4.184(50-20) then the answer would be: 1945.5? lol
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