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

Question

Chemistry

Xander Orr

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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Answers (1)

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Rayne Ray · Answer 1

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.