Using information in Appendices B and C in the textbook, calculate the minimum number of grams of propane, C3H8 (g), that must be combusted to provide the energy necessary to convert 6.15 kg of ice at - 16.5 ∘C to liquid water at 77.0 ∘C.

The minimum number of grams of propane is 84.16 grams

Explanation:

The strategy here is to calculate first the amount of heat required to bring the ice at - 16.5 ºC to water at 77.0 ºC. After this we can calculate the minimum amount of propane which when combusted will give us the heat calculated in the first part.

For the first part we need to know the specific heats of ice and water. The heat of melting will also be needed since the ice needs to be converted to liquid water (phase change).

C (ice) : 2.108 kJ/KgK

C (water) : 4.184 kJ/KgK

Q to bring ice to 0ºC = mxC (ice) xΔT

Q to 0ºC = 6.15 kg x 2.108 kJ/KgK x 16.5 K = 213.9 kJ

Cmelting : 333.55 kJ/Kg

Qmelting = Cmelting x mass

Qmelting = 333.55 kJ/Kg * 6.15 Kg = 2051.33 kK

Q to bring to 77 ºC = m x C (water) xΔT

Q = 6.15 Kg x 4.184 kJ/KgK x (77 K) = 1981.33 kJ

adding the heats calculated to bring the ice @-16.5 ºC to 77 ºC = (213.9+2051.33 + 1981.33) kJ

= 4246.6 kJ

Amount of Propane Required

ΔH⁰combustion propane = - 2220 kJ/mol

therefore for 4246.6kJ we will require

4246.6 kJ x 1 mol/2220 kJ = 1.91 mol

MW propane = 44 g/mol

minimum mass of propane rqd = 1.91 mol x 44 g/mol = 84.2 g