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26 August, 12:05

Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes according to the equation: CO2 (s) →CO2 (g). When dry ice is added to warm water, heat from the water causes the dry ice to sublime more quickly. The evaporating carbon dioxide produces a dense fog often used to create special effects. In a simple dry ice fog machine, dry ice is added to warm water in a Styrofoam cooler. The dry ice produces fog until it evaporates away, or until the water gets too cold to sublime the dry ice quickly enough. Suppose that a small Styrofoam cooler holds 15.0 L of water heated to 90 ∘C. Calculate the mass of dry ice that should be added to the water so that the dry ice completely sublimes away when the water reaches 28?

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  1. 26 August, 13:33
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    6.82 kg

    Explanation:

    Given that the amount of water is 15L and we know that the density of water is ≈ 1kg/L. The mass of water is given by mass = volume x density, i. e,

    mass = 15 x 1 = 15 kg. Also the specific heat capacity of water is 4.186 KJ/kg.

    The sublimation enthalpy of dry ice is 571 KJ/kg.

    Now, the amount of heat lost by water is entirely used up for the sublimation (conversion from soild to gas) of dry ice. And the heat (Q) lost by water is given as : Q = mCΔT, where m is the mass of water, C the specific heat capacity of water and ΔT the change in temperature.

    Here, Q = 15 x 4.186 x (90 - 28) = 3892.98 KJ.

    This amount of heat is taken up by the dry ice for its sublimation. Also the energy taken by dry ice (Q') for its sublimation is given by: Q' = m'L', where m' is the mass of dry ice, L' is the latent heat of sublimation (i. e, the amount of heat required per kg of a substance to sublime) of dry ice amd L' = 571 KJ/kg.

    Now, Q' = m'L' = heat lost by water = 3892.98KJ.

    And, m'L' = m' x 571 KJ/kg = 3892.98 KJ. (Dividing with 571)

    Therefore, m' = 6.82 kg.
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