How many joules are needed to warm 45.0 grams of water from 30.0 degrees C to 75.0 degrees C?

Q=m x Cp x ∆T

m is the mass of the water, Cp is the specific heat of water and ∆T is the change in temperature of the water (final-initial temperature). q is the energy involved in the reaction, measured in joules.

q = (45.0) x (4.184 Jg^-1/°C^-1) x (45°C)

q=8472.6 Joules