What is the energy loss per second for a wall with two different materials?

[dalitwil]

A wall is composed of two materials. Material 1 has a thickness of 7.21 cm and a thermal conductivity of 0.1 while material 2 has a thickness of 6.57 cm and a conductivity of 1. If the temperature difference inside to outside is 25 Co and the wall has an area of 10 m2, what is the energy loss per second to the nearest watt?

My work:
Q1=kA (deltaT/L)t
=(.1)(10)(25/0.0721)t
=346.7t

Q2=kA (deltaT/L)t
=(1)(10)(25/0.0657)t
=3805.2

Qtotal = Q1+Q2
= 346+3805
=4151W
The answer is 318. Any help would be greatly appreciated.

 

[dalitwil]

I think I found the equation i was supposed to use:

Q/t=(A*delta T)/(L/k)

But it still didn't work out. :yuck: Please help!

 

[FredGarvin]

Make sure your units are correct. I am going to have to assume which ones you are using since you do not state them.

First, find the equivilent thermal conduction resistance, U by using:
$$U=\frac{1}{\frac{L_A}{K_A} + \frac{L_B}{K_B}}$$

$$U= \frac{1}{\frac{.0721}{0.1} + \frac{.0657}{1.0}}$$

$$U= 1.271 \frac{W}{m^2 * C}$$

Now use Newton's Law of Cooling in the form of:

$$q_x=UA \Delta T$$

$$q_x=(1.271 \frac{W}{m^2 *C})(10 m^2)(25 C)$$

$$q_x = 317.8 W = 318 W$$