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

"hydrogen bonding is the attraction of h atoms in the water molecule (slight + charge) to o atoms in a neighboring water molecule (slight - charge); it is a weak interatomic force. but it is still strong enough to the water molecules "stick" together."