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What is the specific heat of a substance if 300 J are required to raise the temperature of a 267–g sample by 12ºC? (Must show all work and include units of measure to receive full credit.)

Respuesta :

We can use the heat equation,
Q = mcΔT 



Where Q is the amount of energy transferred (J), m is the mass of the substance (kg), is the specific heat (J g
⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).

 

According to the given data,

Q = 300 J 

m = 267 g

c = ?
ΔT = 12 °C

 

By applying the formula,

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.