Question

How many joules of heat are absorbed to raise the temperature of 435 grams of water at 1 atm from 25°C to its boiling point, 100°C? (Cp of water = 4.18 J/g°C)

Answer 1

The energy absorbed may be calculated using:
Q = mcΔT
Where Q is the energy absorbed, m is the mass of water, c is the specific heat capacity of the water and ΔT is the change is the temperature of the water. Substituting the values,
Q = (435)(4.18)(100-25)
Q = 136,372 J

The energy absorbed is 136.4 kJ