Question

How many Joules of heat are required to heat 75g of ice from -15℃ to 235℃?

-q=mCpΔT
- q=mΔH
- q=mCpΔT
- q=mΔH
- q=mCpΔT
-Total Energy Joules=
-Total energy in kilojoules 1000 J = 1 kJ

Answer 1

78000 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)

Math

Pre-Algebra

Order of Operations: BPEMDAS

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

• Left to Right

Step-by-step explanation:

Step 1: Define

Identify variables

[Given] m = 75 g

[Given] c = 4.184 J/g °C

[Given] ΔT = 235 °C - -15°C = 250 °C

[Solve] q

Step 2: Solve for q

1. Substitute in variables [Specific Heat Formula]: q = (75 g)(4.184 J/g °C)(250 °C)
2. Multiply [Cancel out units]: q = (313.8 J/°C)(250 °C)
3. Multiply [Cancel out units]: q = 78450 J

Step 3: Check

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

78450 J ≈ 78000 J