sue asked in Science & MathematicsPhysics · 4 months ago

entropy changes?

consider the freezing of 1 kg of supercooled water at –10 °C. This is an irreversible

transformation of the water from an initial state (A, supercooled liquid at –10 °C) to a final state (D, ice at –10 °C). The heat of the spontaneous transformation A->D is –315Jg^–1.

Q1

What is the entropy change of the surroundings when 1 kg of supercooled water

freezes at –10 °C? [You may assume that the surroundings represent a large heat

bath, which absorbs and releases heat quasi-statically at –10 °C.]

Q2

Calculate the entropy change of 1 kg of water when it goes along a quasi-static

(“reversible”) path from State A to State D, passing through two intermediate states [B, water at 0 °C; and C, ice at 0 °C). [You may assume that the specific heat of supercooled water is the same as that of ordinary liquid water.]

thank u

info u may need:

Specific heat of water (temperature independent) cwater 4186 J kg^–1 K^–1

Latent heat of freezing of water (at 0 °C) Lf 333 J g^–1

Specific heat of ice (temperature independent) cice 2093 J kg^–1 K^–1

1 Answer

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  • NCS
    Lv 7
    4 months ago
    Favourite answer

    Q1/ entropy change of surroundings

    ΔS = -ΔQ / T = -(-315J/g * 1000g) / 263K

    ΔS = 1198 J/K

    Q2/ consider the reversible path

    (i) heat ice to 0ºC

    ΔS = 1kg * 2093J/kg·ºC * -10ºC  / 263K = -79.6 J/K

    (ii) melt ice at STP

    ΔS = -1kg * 333000J/kg / 273K = -1220 J/K

    (iii) cool water to -10ºC

    ΔS = 1kg * 4186J/kg·ºC * 10ºC / 263K = 159 J/K

    total Q = -312070 J = -3.12e5 J

    ΔS = total Q / T = -3.12e5J / 263K = -1140 J/K

    Questions welcome.

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