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Lubov Fominskaja [6]
2 years ago
12

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 ca

lorimeter 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.
Chemistry
1 answer:
Phoenix [80]2 years ago
6 0

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:

brainly.com/question/6363778

#SPJ1

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Calculate the final molarity of HCl the resulting solution when 5.56 mL of 4.108 M HCl is added to 4.44 mL of water.
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We'll begin by listing what was given from the question. This is shown below:

Initial volume (V₁) = 5.56 mL

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Final volume (V₂) = 5.56 + 4.44 = 10 mL

<h3>Final Molarity (M₂) =? </h3>

The final molarity of the HCl solution can be obtained by using the dilution formula as illustrated below:

<h3>M₁V₁ = M₂V₂</h3>

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