Heat transfer through brass and rubber

[max1995]

Homework Statement

Two identical brass plates, each 0.5 cm thick, are separated by a rubber spacer of thickness 0.1 cm which has the same cross sectional area. If the outer surfaces of the brass plates are kept at 0 °C and 100 °C respectively, calculate the temperature at each side of the rubber spacer given that the thermal conductivity of brass is 500 times bigger than that of rubber.

Homework Equations

I used dQ/dt= -KA*(change in temperature/distance traveled)
k is thermal conductivity of material

The Attempt at a Solution

I know the rate of heat transfer is the same through the whole thing. K of brass is = to 500K of rubber
and the heat transfer goes from the 100 side to the 0 side.

so I made 3 equations

first one (for 100 degree brass to first side of rubber)

dQ/dt= -500kA*(T₂-100/0.5x10^-2)

Second (for change in temperature over rubber (side one to side two

dQ/dt= -kA*(T₃-T2/0.1x10^-2)

third (second side of rubber to 0 degree brass)

dQ/dt= -500kA*(0-T₃/0.5x10^-2)I put 2 of the equations equal to each other but get silly numbers when I solve them (maybe the equations are wrong? or I am solving them wrong?)

thanks for the help

 

[Chestermiller]

Solve for the 3 temperature differences in terms of dQ/dt , and then add them up.