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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.

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Answer:

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

General Formulas and Concepts:
Thermodynamics

Specific Heat Formula: [tex]\displaystyle q = mc \triangle T[/tex]

  • m is mass (g)
  • c is specific heat capacity (J/g °C)
  • ΔT is the change in temperature

Explanation:

Step 1: Define

Identify variables.

m = 112 g

ΔT = 20.0 °C

q = 1004 J

Step 2: Solve for c

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

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

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Topic: AP Chemistry

Unit: Thermodynamics

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