If 41grams of water at 24°C absorbs 2208 J of heat energy, what will be the final temperature of the water?

The problem can solved using the heat equation which is expressed as:

H = mCΔT

where H is the energy absorbed or released, m is the mass of the substance, C is the specific heat capacity, and ΔT is the change in temperature.

2208 J = 41 g x 4.18 J/g·°C x (T - 24 °C)

T = 36.88 °C