Ineed fast what is the specific heat of a substance if 300 j are required to raise the temperature of a 267–g sample by 12ºc? show work

We can use the heat equation,
Q = mcΔT 

Where, 

Q = the amount of energy transferred (J)

m = the mass of the substance (kg)

c = the specific heat (J g⁻¹ °C⁻¹) 

ΔT = the temperature difference (°C).

According to the given data for the substance,

Q = 300 J 

m = 267 g

c = ?
ΔT = 12 °C

From substitution,

300 J = 267 g x c x 12 °C
       c = 0.0936 J g⁻¹ °C⁻¹

Hence, specific heat of the given substance is 0.0936 J g⁻¹ °C⁻¹.

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