How much gas is required to heat a house

[Dyna]

1. Methane has a heat of combustion of 5.5 x 10^7 J/kJ.
a)How much gas is required to heat a house that requires 2.0 x10^9 kJ for the whoel winter? Assume that the furnace is 80% efficient

3. So i tried dividing using Q=mL
but i don't seem to be getting the right answer (4.5 x 10^5)

I also tried using the efficiency formula but that doesn't make sense =\

There is also part b and c but those are reliant on this answer so I'm trying to focus on it but i simply don't understand what i should be doing

 

[Andrew Mason]

1. Methane has a heat of combustion of 5.5 x 10^7 J/kJ.
a)How much gas is required to heat a house that requires 2.0 x10^9 kJ for the whoel winter? Assume that the furnace is 80% efficient

(NOTE: Your units are wonky. The heat of combustion is: 5.5 x 10^7 J/kilogram)

How many joules of combustion heat do you need (this is where the efficiency comes in).

If you get that, it is a simple matter to figure out how many Kg of methane you will need.

AM

 

[Dyna]

So..the house needs 2.0 x 10^9 kJ...
so thats...( 2 x 10^12 x .8) = e/ m

what woudl E be here - 5.5 x 10 ^7 ?

 

[rock.freak667]

1. Methane has a heat of combustion of 5.5 x 10^7 J/kJ.

I am not sure what unit that is supposed to be. But I'd start by finding the energy input using the formula for efficiency.

EDIT: well I was beaten to it, but Andrew Mason has the idea.

So..the house needs 2.0 x 10^9 kJ...

Right, so that is how much energy the methane outputs

 

[Dyna]

Efficiency = useful output / input x 100%
.8 = 2.0 x 10 ^12 J / input x 100%

no i have no idea what I am doing XD help?

 

[rock.freak667]

Efficiency = useful output / input x 100%
80 = 2.0 x 10 ^12 J / input x 100%

Find the input energy perhaps.