Question

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A 74.0-gram piece of metal at 94.0 °C is placed in 120.0 g of water in a calorimeter at 26.5 °C. The final temperature in the calorimeter is 32.0 °C. Determine the specific heat of the metal. Show your work by listing various steps, and explain how the law of conservation of energy applies to this situation.

Answer 1

The specific heat of the metal, given the data from the question is 0.60 J/gºC

Data obtained from the question

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) =?

How to determine the specific heat capacity of the metal

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

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