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A 2.74-g sample of a substance suspected of being pure gold is warmed to 72.1 oC and submerged into 15.2 g of water initially at 24.7 oC. The final temperature of the mixture is 26.3 oC. What is the heat capacity of the substance? Could the substance be pure gold?

(The specific heat capacity of water is 4.184 J/g oC)

J/g oC (Round to three significant digits)

Look up the specific heat capacity of gold.

Could the unknown substance be pure gold? Write yes or no in the following box:

Respuesta :

Eduard22sly

1. The specific heat capacity of the sample of gold obtained is 0.81 J/gºC

2. The substance is not pure gold

1. How to determine the specific heat capacity of gold

  • From the question given above, the following data were obtained:
  • Mass of gold (M₉) = 2.74 g
  • Temperature of gold (T₉) = 72.1 °C
  • Mass of water (Mᵥᵥ) = 15.2 g
  • Temperature of water (Tᵥᵥ) = 24.7 °C
  • Equilibrium temperature (Tₑ) = 26.3 °C
  • Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC
  • Specific heat capacity of gold (C₉) =?

The specific heat capacity of the sample of gold can be obtained as follow:

Heat loss = Heat gain

M₉C₉(M₉ –Tₑ) = MᵥᵥC(Tₑ – Tᵥᵥ)

2.74 × C₉(72.1 – 26.3) = 15.2 × 4.184 (26.3 – 24.7)

C₉ × 125.492 = 101.75488

Divide both side by 125.492

C₉ = 101.75488 / 125.492

C₉ = 0.81 J/gºC

2. How to determine if the substance is pure

  • Specific heat capacity of gold calculated = 0.81 J/gºC
  • Specific heat capacity of pure gold = 0.13 J/gºC

From the above we can conclude that the substance is not pure gold since the calculated and standard value of the specific heat capacity of the gold sample is not the same

Learn more about heat transfer:

https://brainly.com/question/6363778

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