Question

PLEASE HELP ME!!!

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

Answer 1

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

Step-by-step 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