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How much heat must be added to heat 0.750 g of water from 20.0 degrees * C to 40.0C (Remember, Q=C*m* Delta T) The specific heat of water is joule/gram.

How much heat must be added to heat 0750 g of water from 200 degrees C to 400C Remember QCm Delta T The specific heat of water is joulegram class=

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Answer:

D. 62.8 J

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Chemistry

Thermodynamics

Specific Heat Formula: q = mcΔT

  • q is heat (in joules)
  • m is mass (in grams)
  • c is specific heat (in J/g °C or J/g K)
  • ΔT is change in temperature, final minus initial (°C or K)

Explanation:

Step 1: Define

[Given] m = 0.750 g

[Given] c = 4.186 J/g °C

[Given] ΔT = 40.0 °C - 20.0 °C = 20.0 °C

[Solve] q

Step 2: Solve

  1. Substitute in variables [Specific Heat Formula]:                                             q = (0.750 g)(4.186 J/g °C)(20.0 °C)
  2. [Heat] Multiply [Cancel out units]:                                                                    q = 62.79 J

Step 3: Check

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

62.79 J ≈ 62.8 J