Answer:
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
