physics problem conduction?

Question

Three building materials, plasterboard (k=0.30 J/ s.m. C degrees), brick (k=0.60 J/ s.m. C degrees), and wood (k=0.10 J/ s.m. C degrees) are sandwiched together. The temperatures at the inside surface is 27 degrees celcius and 0 degrees celcius on the outside. each material has the same thickness and cross sectional area. Find the temperature
(a) at the plasterboard brick interface and
(b) at the brick wood interface

i'm not sure how to work this problem...

Answer 1

when the steady-state is reached then heat flowing across all intermittent junctions is same or Q-dot =same
Q-dot = - k A dT/dx = - dT/(x/kA)
- ve sign tells that fall in temp and dx are in opposite direction. This like current (motive is potential diff) anology driven here by temp difference (motive force - thermal causality)
I (current) = V/R >>> wherein
x/kA = thermal resistance
Q-dot (0r I same) series circuit
R = R1+R2+R3
R =(x /kpA)+(x /kbA)+(x /kwA)=(x/A)[1/kp+1/kb+1/kw] = 3x/K.A (right side composite slab of 3x thick, K conductivity)
as thickness (x) and A crosstion are same, the joint slab (3x) thickness and A crosssection, joint K of composite slab
3/K = 1/0.3 + 1/0.6 + 1/0.1= 15 >>
K = 0.2 composite
------------
let plaster-brick junction temp = Tpb
let brick-wood interface temp = Tbw
-----------------------
conduction >> Q-dot = - k A dT/dx
Q(27/0) = Q(27/Tpb) = Q(Tbw/0) steady state
-------------------
Q(27/0) = Q(27/Tpb)
(K A/3x) (27-0) = (kp A/x) (27-Tpb)
(K/3) (27-0) = (kp) (27-Tpb)
(0.2/3) (27) = (0.3) (27-Tpb)
(2) (27) = 9 (27-Tpb) =54
Tpb = 21 deg C
------------------------------...
Q(27/0) = Q(Tbw/0)
(0.2/3) (27) = (0.1) (Tbw - 0)
Tbw = 18 deg C
temps are
27 ---- 21 ----- 18 ----- 0
outside ---- Tpb ----- Tbw ----- inside
.........plaster .... brick ...... wood .....

Answer 2

It's a classic heat flow problem. Look in your book and write an equation for the flow of heat through the 3 layers with the delta T being 27 degrees. They do not talk about equilibrium, but they undoubtedly mean to ask what the temperatures are at equilibrium, since the temperature will change with time until equilibrium is reached.

Answer 3

Think of it as three series resistors and an applied voltage. Does that work for you?