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If 2050 J of heat are added to a 150 g object its temperature increases by 15°C.
(a) What is the heat capacity of this object?
(b) What is the object's specific heat?

Respuesta :

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When an object gets heated by a temperature ΔT energy needed, E = mcΔT

Here energy is given E = 2050 J

Mass of object = 150 g

Change in temperature ΔT  = 15 [tex]^0C[/tex] = 15 K

a) Heat capacity of an object equal to the ratio of the heat added to (or removed from) an object to the resulting temperature change.

  So heat capacity = E/ΔT = 2050/15 = 136.67 J/K

b) We have E = mcΔT

                    c = [tex]\frac{2050}{150*10^{-3}*15}  = 911.11 J/kgK[/tex]

 So object's specific heat = 911.11 J/kgK

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