What is the final temperature after 840 J is absorbed by 10.0 g of water at 25°C?

1 Answer
Dang Thi ThanhHoa · Grace
May 15, 2016

The final temperature is equal to 45^@45∘ CC

Explanation:

We have the amount of energy gain
Q=m*c*DeltaT=mcDeltat

where c = 4.184 J/g.C is the specific heat of water, m is the mass of water

=>840=10 x 4.184*(t-25)
t=840/(10" x "4.184)+25=45 i.e. 45^@ C

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