Objective:The objective of this experiment was to demonstrate that different types of materials (in our case brass, aluminum, and stainless steel) have different rates of heat conduction.Introduction:Heat conduction plays an important role in the everyday functions of a building. For example, it is a crucial part of the fire-tube boilers, which are commonly used to heat residential and commercial buildings. Heat conduction, in the general sense, is a transfer of heat from a surface of higher temperature to lower temperature. This experiment will test the rate of heat conduction of three different materials (two brass, one aluminum, and one stainless steel) of the same size, except for brass which will have two different widths. Each of the four pieces material will contain two thermocouples, labeled from T1 to T8.MaterialDimensionThermocoupleBrass12mm x 4.2mmT1,T2Brass08mm x 4.2mmT3, T4Aluminum12mm x 4.2mmT5,T6Stainless Steel12mm x 4.2mmT7,T8Table 1The ends of each piece of material (closest to the center) will be heated, as heat flows through the pieces of material, the temperature will be recorded by each thermocouple. This will demonstrate that different thermal conductivity’s and different material sizes will influence the rate at which heat is being transferred for a specific time. Fourier’s law of heat conduction allows the rate of heat transfer to be found through a section of thickness.Q=KAΔTΔx(1)Where: Q is rate of heat transfer per unit time [J/s]K is the thermal conductivity coefficient (varies for different materials) [W/m°C]A is the cross sectional area (perpendicular to the direction of transfer) [m2]Δx is length of the material [m]ΔT is the difference in temperature (Tf– Tc). Where Tfrepresent the temperature at a point further away from heat source and Tcis closest to the heat source. [°C]The rate of heat transfer per unit time can also be calculated by:

Q=VI(2) Where: Q is rate of heat transfer per unit time [J/s]V is the input voltage [v]I is the amount of current supplied [A]Substituting equation (1) into (2) KAΔTΔx=VI(3)We can then solve for the thermal conductivity coefficientK=VIΔxAΔT(4)If area and distance between the thermocouples are held constant, then the difference in