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 your work

               Cp  =  0.093 J.g⁻¹.°C⁻¹
Solution:

The equation used for this problem is as follow,

                                                  Q  =  m Cp ΔT    (1)

Where;
            Q  =  Heat  =  300 J

            m  =  mass  =  267 g

            Cp  =  Specific Heat Capacity  =  ??

            ΔT  =  Change in Temperature  =  12 °C

Solving eq. 1 for Cp,

                                 Cp  =  Q / m ΔT

Putting values,
                                 Cp  =  300 J / (267 g × 12 °C)

                                 Cp  =  0.093 J.g⁻¹.°C⁻¹

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