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