Ask Question
25 May, 17:50

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)

+4
Answers (1)
  1. 25 May, 21:11
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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 ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers