Answer:
62.7KJ
Explanation:
How many joules are required to heat 200 grams of water from 25˚C to 125˚C? 524.8 kJ
Start with Specific Heat because the water is not going through a phase change.
m= 200 g
Cp= 4.18 J/g˚C .......must be given
∆T= 100˚C - 25˚C = 75˚C
q =mCp∆T
q = (200g)(4.18 J/g˚C)(75˚C)
q = 62700 J = 62.7 kJ
