What is the enthalpy change during the process in which 100.0 g of water at 50.0 °C is cooled to ice at - 30.0 °C?

(Use the specific heats and enthalpies for phase changes)

60.1 kJ

Explanation:

Enthalpy change ΔH = m x Δt x cρ

Given that:

100 grams water = 5.56 moles water

The enthalpy change needed to raise 100 g water from 50°C to 0°C is calculated as:

ΔH = 100g x (0-50) x 4.18J/gC = - 2.09 X 10⁴ Joules

The Freezing water

ΔH = ΔH fusion * mol

ΔH fus for water is = 6.01 kJ/mol

ΔH = 6.01 kJ/mol * 5.56 moles = - 33.4 kJ (since heat is released when water freezes)

Finally, The enthalpy Change during the process pf changing the ice from 0°C to - 30.0°C is:

ΔH = m x Δt x cp

= 0.100 kg * (-30 - 0) °C * 2.00 * 10³ J/kgC

= - 6 * 10³ J

Total heat lost = - 2.09 * 10⁴ J + (-33.4 * 10³ J) + (-6 * 10^3J)

= - 6.01 * 10⁴ J

= 60.1 kJ