How many kilograms of propane would be needed to ....? IM STUCK ON THIS PROBLEM! HELP :( :(?

Question

A house with a 16.2 × 103 ft2 floor area and 8.50 ft high ceilings is to be heated with propane. How many kilograms of propane would be needed to raise the temperature of the air inside the house from 63.0 oF to
73.0 oF ?. In this calculation your are ignoring the temperature of the walls etc.



The moloar specific heat of the air inside the house is 30.2 J mol −1 oC −1
The average density of air over this temperature range can be estimated to be 1.22 g/L.
The molar mass of air is 29.1 g/mol.
There are 28.317 liters in a cubic foot.
The molar heat of combustion of propane is −2220.047 kJ/mol


PLEAASE HELPPP I NEED ANSWER + EXPLANATION IF POSSIBLE :) thankyoooou!

Answer 1

Assuming your area translates to 16200 ft^2 and not 16.2 ft x 1000 ft. So righty o then.....

Volume = 16200 x 8.5 = 137700 ft^3 = 137700 x 28.317 = 3.899 x 10^6 L

Mass = 3.899 x 10^6 x 1.22 = 4.757 x 10^6 g = 4.757 x10^6/29.1=1.635 x 10^5 mol

Temp Rise = 10 oF = 5.56 oC

Specific Heat od Air = 30.2 J/mol/oC

Enthalpy = 1.635 x 10^5 x 30.2 x 5.56 / 1000 = 27454 kJ

Heat of Combustion C3H8 = 2220 kJ/mol

C3H8 mol = 27454 / 2220 = 12.37 mol

MW C3H8 = 3 x 12 + 8 = 44

C3H8 kg = 12.37 x 44 / 1000 = 0.544 kg

So follow the maths and make sure I got it right and change the area if thats wrong.

Onya.....

Answer 2

I went through this same problem for another questioner last night, but I'll do it again. It's lots of arithmetic, but nothing conceptually difficult. Find:
- The volume of the house, in cubic feet.
- Volume in liters.
- Mass of the air, in grams.
- Number of moles of air.
- Heat required (in joules), using the specific heat and the temperature change. (Convert from F to C.)
- Number of moles of propane required.
- Weight of propane, using its molecular weight (formula C3H8).
- Something cold to drink.

Answer 3

WOW ! Thats one dam interesting question , I wish I could answer it for you.