How much energy must be absorbed by water with a mass of 0.5 kg in order to raise the tempature from 30 degrees celcius to 65 degrees celcius?

Q = 73.173 KJ

Explanation:

Q = m*C*ΔT

∴ C: water heat capacity

∴ C H2O (25°C) = 4.1813 J/g. K

∴ m H2O = 0.5 Kg = 500 g

∴ T1 = 30°C ≅ 303 K

∴ T2 = 65°C ≅ 338 K

⇒ ΔT = 338 - 303 = 35 K

⇒ Q = (500 g) * (4.1813 J/g. K) * (35 K)

⇒ Q = 73172.75 J = 73.173 KJ