Question

A 76.0-gram piece of metal at 96.0 °C is placed in 120.0 g of water in a calorimeter at 24.5 °C. The final temperature in the calorimeter is 31.0 °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 capacity of the metal given the data from the question is 0.66 J/gºC

Data obtained from the question

• 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) =?

How to determine the specific heat capacity of the metal

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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