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A 148.20 sample of an unknown substance was heated from 25°C to 40°C. In the process, the substance absorbed 5683 J of energy. What is the
specific heat of the substance.

Respuesta :

Answer: 2556.455 J/(kg.°C)

Explanation:

Im assuming the 148.20 is in grams

The equation is Q=mcΔT

Q= Heat (J)

m= Mass

c= SHC (specific heat capacity)

ΔT= Temp change

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