- #1
max1995
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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