What is the specific heat of a substance if 373 J is required to raise the temperature of a 312 g sample by 15°C?

0.080 J/g °C

General Formulas and Concepts:

Thermodynamics

Specific Heat Formula: q = mcΔT

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

Explanation:

Step 1: Define

[Given] q = 373 J

[Given] m = 312 g

[Given] ΔT = 15 °C

[Solve] c

Step 2: Find Specific Heat

Substitute in variables [Specific Heat Formula]:                                        373 J = (312 g)c(15 °C)Multiply:                                                                                                        373 J = (4680 g °C)cIsolate c:                                                                                                        c = 0.079701 J/g °C

Step 3: Check

Follow sig fig rules and round. We are given 2 sig figs as our lowest.

0.079701 J/g °C ≈ 0.080 J/g °C

Topic: AP Chemistry

Unit: Thermodynamics

Copper, silver, aluminum, cobalt, chromium, mercury 

The answers are 3.c kelvin4.b liquids only