Calculating Heat Flow in Composite Sheet with Different Thermal Conductivities

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To calculate heat flow in a composite sheet with different thermal conductivities, first determine the thermal resistance of each layer, treating them as resistances in series. The formula for heat transfer can be adapted from Ohm's law, where the total temperature difference is equal to the sum of the thermal resistances multiplied by the heat flow. Use the individual thermal conductivities of rubber and cardboard to find their respective resistances based on their thickness and area. Finally, apply the total thermal resistance to calculate the heat flow across the composite sheet over the specified time. Understanding this method will clarify how to approach problems involving multiple materials with varying thermal conductivities.
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1. A sheet of rubber and a sheet of cardboard, each 2 mm thick, are pressed together and their outer faces are maintained respectively at 0 oC and 25 oC. If the thermal conductivities of rubber and cardboard are respectively 0.13 and 0.05 W m-1 K-1, find the quantity of heat which flows in one hour across a piece of the composite sheet of area 100 cm2. 2. I know how to solve the problem when a single object with one thermal conductivity is give, But how do I solve a problem with two different objects with two different thermal conductivities?3. I used the equation Heat conduction Q/ Time = (Thermal conductivity) x (Area) x (Thot -Tcold)/Thickness to find the Heat which flows in the rubber and the cardboard seperately but after that I was at a lost.

Please help me solve it. An explanation would be nice if possible because I don't really care about the answer I just want to know how to do it.
Thanks in Advance :)
 
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Each of the layers can also be seen as a thermal resistance (which is the inverse value of the conductivity). Accoring to Ohm's law (which is much more common: U = R⋅I), you can also establish the same formula for heat transfer/conduction: ΔT = RTh ⋅ Q. You can find the (total) resistance of thermal resistances in series and parallel like it is done with electrical resistances/impedances. It#s up to you if you transform the gained RTh back into a conductivity and use Q = K ⋅ ΔT or Q = ΔT / RTh ⋅A. Just be careful with the dimenions of K and RTh, but as you know the formula, the adaption should be easy.
 
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