The specific heat of the metal, given the data from the question is 0.60 J/gºC
<h3>Data obtained from the question </h3>
The following data were obtained from the question:
- Mass of metal (M) = 74 g
- Temperature of metal (T) = 94 °C
- Mass of water (Mᵥᵥ) = 120 g
- Temperature of water (Tᵥᵥ) = 26.5 °C
- Equilibrium temperature (Tₑ) = 32 °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 gold can be obtained as follow:
According to the law of conservation of energy, we have:
Heat loss = Heat gain
MC(T –Tₑ) = MᵥᵥC(Tₑ – Tᵥᵥ)
74 × C(94 – 32) = 120 × 4.184 (32 – 26.5)
C × 4588 = 2761.44
Divide both side by 4588
C = 2761.44 / 4588
C = 0.60 J/gºC
Thus, the specific heat capacity of the metal is 0.60 J/gºC
Learn more about heat transfer:
brainly.com/question/6363778
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