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The specific heat capacity of the metal given the data from the question is 0.66 J/gºC
The specific heat capacity of the sample of the metal can be obtained as follow:
Heat loss = Heat gain
MC(M –Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)
76 × C × (96 – 31) = 120 × 4.184 × (31 – 24.5)
C × 4940 = 3263.52
Divide both side by 4940
C = 3263.52 / 4940
C = 0.66 J/gºC
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