How much heat is absorbed when 447 g of water , Cp = 4.184 J/goC, goes from 25.0 oC to 35.0 oC?

The answer to your question is  Q = 18702.5 J

Explanation:

Data

mass of water = m = 447 g

Cp = 4.184 J/g°C

Temperature 1 = T1 = 25°C

Temperature 2 = T2 = 35°C

Heat = Q = ? Joules

Process

1.- Write the formula to calculate heat

    Q = mCp(T2 - T1)

2.- Substitution

    Q = (447)(4.184)(35 - 25)

3.- Simplification

    Q = (447)(4.184)(10)

4.- Result

    Q = 18702.5 J

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8 m/s because for him to get the same amount of energy as ray is putting out, bobby, who weighs half as much as ray, needs to run twice as fast.