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.

0.424 J/g °C

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition 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 temperatureExplanation:

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

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