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.

D. 62.8 J

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Thermodynamics

Specific Heat Formula: q = mcΔT

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

Step 3: Check

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

62.79 J ≈ 62.8 J

that's just how they are

explanation:

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