Question

Propane, C3H8, is used in many instances to produce heat by burning: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) The standard enthalpy of reaction, ΔHrxn, is −2,044 kJ. ΔHrxn is a symbol or the heat of reaction. A swimming pool is 35 meters long, 4 meters deep and 15 meters wide is filled with water. The temperature of the water is 20°C, but the owner of the pool would like the temperature to be 30°C. How many grams of propane would have to be consumed if all of the heat from the combustion was to be absorbed by the water?

Answer 1

Ans. Step 1: Calculate volume of water in swimming pool-

Volume of swimming pool = l x b x depth

                                                = 35.0 m x 15.0 m x 4.0 m

                                                = 2100.0 m³

                                                = 2100.0 x (10³ L)                             ; [1 m³ = 10³ L]

                                                = 2.10 x 10⁵ L

Assuming the pool to be filled with water, volume of water in it = 2.10 x 10⁵ L

Assuming density of water is equal to 1.000 kg/L at given temperature, the masss of water in pool = volume x density

            = 2.10 x 10⁵ L x (1.000 kg/ L)

            = 2.10 x 10⁵ kg

Step 2: Calculating amount of heat required to increase pool’s temperature-

Amount of heat required to warm water in pool, q is given by the equation,

q = m s dT                            - equation 1

Where,

q = heat required

m = mass of water in kg

s = specific heat of water [ s = 4.184 kJ kg^-10C^-1 for water]

dT = Final temperature – Initial temperature = 10.0⁰C

Putting the values in equation 1-

            q = (2.10 x 10⁵ kg) x (4.184 kJ kg^-10C^-1) x 10.0⁰C = 87864000.00 kJ

            Hence, q = 87864000.00 kJ                                 

Step 3: Calculating mass of propane required:

Given, dH_rxn = - 2044 kJ for combustion of 1 mol propane as shown in the balanced reaction. That is, combustion of 1 mol propane releases 2044 kJ energy.

So, enthalpy of combustion = - 2044 kJ/ mol. Note that the –ve sign simply indicates that heat is released during the process.

The total amount of heat required to warm water must be equal to total heat released by combustion of propane.

So, total heat to be released by propane sample = - 87864000.00 kJ

Now,

Moles of propane required = total heat to be released / enthalpy of combustion

                                                = (- 87864000.00 kJ) / (- 2044 kJ/mol)

                                                = 42986.30137 mol

Mass of propane required = Moles x Molar mass

                                                = 42986.30137 mol x (44.09652 g/mol)

                                                = 1895546.30 g