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

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

to RightExplanation:

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

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

Are you going to give an example of the plant or?

Naoh + hbr -> nabr (salt) + h2o ca (oh)2 + 2hbr - - - > ca (br)2 + 2h2o