Answer to Question #116421 in Chemistry for Lauren

Question

The specific heat capacity of water is 4.18 J/(g °C). In a cup, 100 grams of water was put in the microwave to heat it from a temperature of 25 °C, to right at boiling temperature. What is the enthalpy of this water?

Answer 1

Solution:

The energy, Q, needed to increase the temperature of an object of mass m, by a temperature ΔT, where the substance's specific heat is C, is given by:

Q = m × C × ΔT

Here,

m = 100 g

C = 4.18 J/(g^∘C)

ΔT = T_final - T_initial = (100 − 25)^∘C = 75^∘C

Inputting:

Q = (100 g) × (4.18 J/(g^∘C) × (75^∘C) = 31350 J

Q = ΔH

ΔH = 31350 J

Answer: The enthalpy of this water is 31350 Joules.