Question

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.

Answer 1

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

General Formulas and Concepts:
Thermodynamics

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

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

Step-by-step 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]:
   `\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 c:
   `\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 c is equal to around 0.45 J/g °C.

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Topic: AP Chemistry

Unit: Thermodynamics