To find the heat that is required to heat 100 grams of water from 0 °C to 25.0 °C we have to use this equation:
Q = m * c * ΔT
In this equation Q is the heat, m is the mass, c is the specific heat and ΔT is the change in temperatue.
m = mass = 100 g
c = specific heat = 4.18 J/(g °C)
ΔT = Tfinal - Tinitial = 25.0 °C - 0.0 °C = 25.0 °C
ΔT = 25.0 °C
Using the given values we can find the answer to our problem.
Q = m * c * ΔT
Q = 100 g * 4.18 J/(g °C) * 25.0 °C
Q = 10450 J
Answer: 10450 J are required.