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A 34.34 g sample of a substance is initially at 26.7 °C. After absorbing 2205 J of heat, the temperature of the substance is 152.1 °C. What is the specific heat (c) of the substance?

Respuesta :

gilcarusolautaro

Answer:

The specific heat of the substance is c= 512.04 J/kg K

Explanation:

ΔQ= 2205 J

m= 0.03434 kg

ΔT= 125.4 ºC

ΔQ= m * c * ΔT

c= ΔQ / (m * ΔT)

c= 512.04 J/Kg K

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