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

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

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right 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

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

Ok where is the image ?