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gogolik [260]
3 years ago
8

A 112 gram sample of an unknown metal was heated from 0.0C to 20.0C. The sample absorbed 1004 J of energy. What was the specific

heat capacity of this metal? Show your work.
Chemistry
1 answer:
worty [1.4K]3 years ago
5 0

Answer:

The specific heat of the sample unknown metal is approximately 0.45 J/g °C.

General Formulas and Concepts:
<u>Thermodynamics</u>

Specific Heat Formula: \displaystyle q = mc \triangle T

  • <em>m</em> is mass (g)
  • <em>c</em> is specific heat capacity (J/g °C)
  • Δ<em>T</em> is the change in temperature

Explanation:

<u>Step 1: Define</u>

<em>Identify variables.</em>

<em>m</em> = 112 g

Δ<em>T</em> = 20.0 °C

<em>q</em> = 1004 J

<u>Step 2: Solve for </u><u><em>c</em></u>

  1. Substitute in variables [Specific Heat Formula]:                                        \displaystyle 1004 \ \text{J} = (112 \ \text{g})(c)(20.0 \ ^{\circ} \text{C})
  2. Simplify:                                                                                                        \displaystyle 1004 \ \text{J} = (2240 \ \text{g} \ ^\circ \text{C})c
  3. Isolate <em>c</em>:                                                                                                        \displaystyle c = 0.448214 \ \text{J} / \text{g} \ ^\circ \text{C}
  4. Round [Sig Figs]:                                                                                          \displaystyle c \approx 0.45 \ \text{J} / \text{g} \ ^\circ \text{C}

∴ specific heat capacity <em>c</em> is equal to around 0.45 J/g °C.

---

Topic: AP Chemistry

Unit: Thermodynamics

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yaroslaw [1]
<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

  • Mass of the unknown metal sample as 58.932 g
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  • Final temperature of metal is 23.68 °C
  • Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

  • Therefore; mass of pure water is 45.2 g
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We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

                       = 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

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<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q =  m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

                                              = 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

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<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
  • We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
  • Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

  = 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

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To know how many moles reacts we need to find limiting reactant:

Moles HCl = 0.050L ₓ (1.27mol /  L) = 0.0635 moles HCl

Moles NaOH = 0.050L ₓ (1.32mol /  L) = 0.066 moles NaOH

As there are more moles of NaOH than moles of HCl, <em>HCl is limiting reactant and moles of reaction are moles of limiting reactant, </em><em>0.0635 moles</em>

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Using the coffee-cup calorimeter equation we can find how many heat was released thus:

Q = C×m×ΔT

<em>Where Q is heat released, C is specific heat of the solution (4.18J/g°C), m is mass of solution (100g because there are 100mL of solution -50.0mL of HCl and 50.0mL of NaOH- and density is 1g/mL) and ΔT is change in temperature (8.49°C)</em>

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Q = 3548.8J of heat are released in the reaction

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ΔH = -55887J / mol

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