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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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<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
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<u>Step 1: Define</u>

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[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

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  2. [DA} Multiply/Divide [Cancel out units]:                                                         \displaystyle 5.55001 \ mol \ C_2H_5OH

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

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