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4130 J of energy is added to a 52 g sample of water that has an initial temperature of 10.0°C. If the specific heat of water is 4.18 J/(g × °C), what would the final temperature of the water be?

Respuesta :

DeanWinchester16
q=mCΔT
4130=(52)(4.18)(Tf-10)
4130=(217.36)(Tf-10)
(4130/217.36)+10=Tf
Tf=29°C

Answer:

29 c and i got it for sure right on the test

Explanation:

q=mc delta T

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