calculate the minimum number of grams of C3H8(g) that must be combusted to provide the energy necessary to convert 3.36 kg of H2O from its solid form at -10.0°C to its liquid form at 80.0°C

Answer 1

To solve this problem, we are going to find out how much energy is required to convert the ice at -10.0°C to liquid water at 80.0°C in two steps:

1. Heating the ice from -10.0°C to 0.0°C.
2. Melting the ice and subsequently heating the liquid to 80.0°C

For step 1:
Using the heat capacity of ice (Cice) = 2.093 J/g°C, we can calculate the energy required as follows:

q1 = mass * Cice * ΔT
where mass = 3.36 kg = 3360 g, ΔT = 10.0°C
q1 = 3360g * 2.093 J/g°C * 10.0°C
q1 = 70312.8J

For step 2:
Melting the ice and heating the liquid water to 80.0°C.

Melting the ice at 0°C to water will require the energy (q2) as follows:
q2 = mass * Lf
where Lf (latent heat of fusion) = 334 J/g
q2 = 3360g * 334 J/g
q2 = 1122240J

Heating the water from 0°C to 80°C will require energy (q3) as follows:
Using the heat capacity of water (Cwater) = 4.186 J/g°C
q3 = mass * Cwater * ΔT
ΔT = 80°C
q3 = 3360g * 4.186 J/g°C * 80.0°C
q3 = 1123555.2J

Total energy required (q) = q1 + q2 + q3 = 70312.8J + 1122240J + 1123555.2J = 2316107.8 J.

Now we need to find the amount of C3H8 combustion needed to produce that energy. The combustion reaction for propane is:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

1 mole propane combustion gives ΔH = -2220 kJ/mol

Energy produced in the reaction = -ΔH = 2220 kJ/mol

We need to find out how many moles of C3H8 are required to produce 2316107.8 J.

Moles of propane = Energy required / Energy produced per mole
= 2316107.8 J / 2220 kJ/mol
= 2316107.8 J / (2220 * 1000) J/mol
= 1.04260 mol of propane

The molar mass of propane (C3H8) is 3(12.01) + 8(1.008) = 44.096 g/mol.

So the mass of C3H8 required is mass = moles * Molar mass
= 1.04260 mol * 44.096 g/mol
= 45.98 g (approximately)

The minimum number of grams of C3H8(g) that must be combusted to provide the energy necessary to convert 3.36 kg of H2O from its solid form at -10.0°C to its liquid form at 80.0°C is approximately 45.98 g.