The specific heat capacity of the metal given the data from the question is 0.66 J/gºC
<h3>Data obtained from the question</h3>
- Mass of metal (M) = 76 g
- Temperature of metal (T) = 96 °C
- Mass of water (Mᵥᵥ) = 120 g
- Temperature of water (Tᵥᵥ) = 24.5 °C
- Equilibrium temperature (Tₑ) = 31 °C
- Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC
- Specific heat capacity of metal (C) =?
<h3>How to determine the specific heat capacity of the metal</h3>
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
Learn more about heat transfer:
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