5 Thermo 2

# 5 Thermo 2 - Setting: A rectangular slab of thickness x and...

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the area of the slab, the temperature difference, Δ T , between the back and the front; and inversely proportional to Setting: A rectangular slab of thickness Δ x and with an area A . The front side of the slab is at a temperature T ; the back side has a somewhat different temperature, T+ Δ T . We are trying to calculate the heat- flow rate, the amount of heat flowing through the slab per unit time, H = Δ Q/ Δ t . the thickness of the slab, Δ x . H should also somehow depend on properties of the material of the slabE We expect H to be proportional to

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Bringing all the parts together: H = Δ Q/ Δ t – heat-flow rate is measured in Joules/second, J/s , or Watts, W . Thermal conductivity, k , is measured in W/(m K) . x T kA H Δ Δ = The coefficient k reflects specific properties of the material of the slab and is called thermal conductivity
Thermal conductivities of different materials. Best heat conductor – Copper; use it when you build a heat sink, as a material for pipes in your cooling system, a radiator. Worst heat conductors are the best insulating materials – air, fiberglass (layers in the walls of houses in cold regions), styrofoam (cups for your hot coffee).

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Heat-flow rate equation Continuing the analogy: electric resistance, R , is analogous to ) /( kA x T x T kA H Δ Δ = Δ Δ = Is similar to the Ohm’s law: R V I = The current, I = Δ q/ Δ t , amount of charge per unit time,
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## This note was uploaded on 12/13/2011 for the course PHYS 2C PHYS 2C taught by Professor Groisman during the Spring '11 term at UCSD.

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5 Thermo 2 - Setting: A rectangular slab of thickness x and...

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