Question

How much heat is required to raise the temperature of 225 grams of ice from -26.8 °C to steam at 133 °C ?

Answer 1

Answer:

150000 J

General Formulas and Concepts:

Chemistry

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 or K)

Math

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Explanation:

Step 1: Define

Identify variables

[Given] m = 225 g

[Given] c = 4.184 J/g °C

[Given] ΔT = 133 °C - -26.8 °C = 159.8 °C

[Solve] q

Step 2: Solve for q

1. Substitute in variables [Specific Heat Formula]:                                          q = (225 g)(4.184 J/g °C)(159.8 °C)
2. Multiply:                                                                                                           q = (941.4 J/°C)(159.8 °C)
3. Multiply:                                                                                                           q = 150436 J

Step 3: Check

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

150436 J ≈ 150000 J

Topic: AP Chemistry

Unit: Thermodynamics

Book: Pearson AP Chemistry