What is the enthalpy change during the process in which 100.0 g of water at 50.0 C is cooled to ice at -30 oC

33600 J

Explanation:

From the question,

ΔH = cm(t₂-t₁) Equation 1

Where, ΔH = Enthalpy change of water, c = specific heat capacity of water, m = mass of water, t₂ = Final temperature of water, t₁ = Initial temperature of water

Given: c = 4200 J/kg.°C, m = 100 g = 0.1 kg, t₂ = 50°C, t₁ = -30°C

Substitute these values into equation 1

ΔH = 4200(0.1)(50+30)

ΔH = 420(80)

ΔH = 33600 J