Question

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

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.