Question

Propane, C3H8, is used in many instances to produce heat by burning:

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

The standard enthalpy of reaction, ΔHrxn, is −2,044 kJ.

A swimming pool is 9.0 meters long, 1.0 meters deep and 3.0 meters wide is filled with water. The temperature of the water is 17°C, but the owner of the pool would like the temperature to be much warmer. The pool owner has a 120 gallon propane tank filled with 174 L of propane at 14.0 atm and 302 K. If the owner burns all of the propane in the tank, and assuming all of the heat goes to warming pool, what is the final temperature of the pool? The specific heat of water is 4.184 J/g · °C.

Answer 1

We have to calculate the moles of propane that will be consumed, the mass of the water in the pool and finally the temperature asked

moles of propaine to be consumed with ideal gas equation

PV=nRT, p is pressure, v is volume, n is moles, R gas constant and T temperature , this ones are given in the statement

14 * 174 =n * 0.82 * 302

n = 14 * 174 / (0.082 * 302) = 98.36 moles of propane are available

From the statement we know that for every mole of propane consumed 2044 KJ are released

so total heat from the actual ammount of propane 98.36 * 2044 = 201 047.84 KJ

For the mass of the water calculate the volume, just multiply the measures

deep * wide * long = 9 * 1 * 3 = 27 m³

density of water is 1000 kg / m³

density = mass / volume

mass = density * volume = 1000 kg / m3 * 27 m3 = 27 000 Kg of water

now we just need to apply the equation

Enthalpy = mass * heat capacity * (Tf - Ti), Tf and Ti are final and initial temperatures

heat capacity is 4.184 J / g which is the same to say 4.184 KJ / Kg

201 047.84 = 27 000 * 4.184 * (Tf - 17), the positive value is because the water is absorbing heat

201 047.84 / (27 000 * 4.184) = 1.78

Tf - Ti = 1.78

Tf = Ti + 1.78 = 17 + 1.78 = 18.78 C

*please rate this answer if you like it =)