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A 38.8 g piece of metal alloy absorbs 181 J as its temperature increases from 25.0°C to 36.0°C. What is the alloy's specific heat?

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Answer :

Answer:

0.424 J/g °C

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality

Chemistry

Thermochemistry

Specific Heat Formula: q = mcΔT

  • q is heat (in Joules)
  • m is mass (in grams)
  • c is specific heat (in J/g °C)
  • ΔT is change in temperature

Explanation:

Step 1: Define

[Given] m = 38.8 g

[Given] q = 181 J

[Given] ΔT = 36.0 °C - 25.0 °C = 11.0 °C

[Solve] c

Step 2: Solve for Specific Heat

  1. Substitute in variables [Specific Heat Formula]: 181 J = (38.8 g)c(11.0 °C)
  2. Multiply: 181 J = (426.8 g °C)c
  3. [Division Property of Equality] Isolate c: 0.424086 J/g °C = c
  4. Rewrite: c = 0.424086 J/g °C

Step 3: Check

Follow sig fig rules and round. We are given 3 sig figs.

0.424086 J/g °C ≈ 0.424 J/g °C

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Rewritten by : Barada