Conductance of heat through concrete

[slaw155]

Homework Statement

In an electrically heated home, the temperature of the ground in contact with a concrete basement wall is 10.7 oC. The temperature at the inside surface of the wall is 21.4 oC. The wall is 0.17 m thick and has an area of 6.4 m2. Assume that one kilowatt hour of electrical energy costs $0.10. How many hours are required for one dollar's worth of energy to be conducted through the wall?

Homework Equations

heat = (thermal conductivity constant x change in temp x area x time)/length

The Attempt at a Solution

heat energy to be conducted = 3.6 x10^6J x 10
(1.1 x (21.4-10.1) x 6.4t)/0.17 = 443 x time
so equating these gives 3.6 x 10^7 = 443t
however this t value is much too large, where have I gone wrong?

 

[NascentOxygen]

So, you calculated how many hours, exactly?

 

[rude man]

Homework Statement

In an electrically heated home, the temperature of the ground in contact with a concrete basement wall is 10.7 oC. The temperature at the inside surface of the wall is 21.4 oC. The wall is 0.17 m thick and has an area of 6.4 m2. Assume that one kilowatt hour of electrical energy costs $0.10. How many hours are required for one dollar's worth of energy to be conducted through the wall?

Homework Equations

heat = (thermal conductivity constant x change in temp x area x time)/length

right so far.

so what is the thermal conductivity constant?

The Attempt at a Solution

heat energy to be conducted = 3.6 x10^6J x 10
(1.1 x (21.4-10.1) x 6.4t)/0.17 = 443 x time
so equating these gives 3.6 x 10^7 = 443t
however this t value is much too large, where have I gone wrong?

where did the numbers "3.6e6J" and "1.1" come from? Is "10" the number of kwh? kw-h is not an SI unit.

use the thermal conductivity number and change total energy from kw-h to J.