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23 November, 06:54

using information in titles of appendices b and c, calculate the minimum grams of propane, C3H8 (g), that must be combusted to provide the energy necessary to convert 5.50 kg of ice at - 20C to liquid water at 75C

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  1. 23 November, 09:54
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    Given:

    Mass of ice = mass of water = 5.50 kg = 5500 g

    Temperature of ice = - 20 C

    Temperature of water = 75 C

    To determine:

    Mass of propane required

    Explanation:

    Heat required to change from ice to water under the specified conditions is:-

    q = q (-20 C to 0 C) + q (fusion) + q (0 C to 75 C)

    = m*c (ice) * ΔT (ice) + m*ΔHfusion + m*c (water) * ΔT (water)

    = 5500[2.10 (0 - (-20)) + 334 + 4.18 (75-0) ] = 3792 kJ

    The enthalpy change for the combustion of propane is - 2220 kJ/mol

    Therefore, the number of moles of propane corresponding to the required energy of 3792 kJ = 1 mole * 3792 kJ/2220 kJ = 1.708 moles of propane

    Molar mass of propane = 44 g/mol

    Mass of propane required = 1.708 moles * 44 g/mol = 75.15 g

    Ans: 75.15 grams of propane must be combusted.
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