Question

Considering this analogy (current = rate that heat flows through a window, and voltage = temperature difference across the window), how does the resistance of an object depend on the length of the object through which current flows? (This is equivalent to the relationship between the rate of thermal energy transfer and the thickness of the window through which this energy flows.)

Answer 1

Answer:

Explained

Explanation:

Resistance R in a current flow through an object is given by

$R= \frac{\rho L}{A}$

ρ = resistivity of the material

L= length of the object

A=  area of cross section

clearly resistance is directly dependent on length of the object.This means greater the length larger will be resistance to current.

thermal resistance R_th is given by

$R_{th} = \frac{L}{KA}$

L= length of the object

A=  area of cross section

K = Conductivity of the material

thermal resistance is also is directly dependent on length of the object.This means greater the length larger will be resistance to current.

Answer 2

Answer:

$\dot Q\equiv I$

$R\equiv \frac{dx}{k.A}$

$\Delta T\equiv V$

$\dot Q=\frac{\Delta T}{(\frac{dx}{k.A} )}$

$I=\frac{V}{R}$

Explanation:

For heat transfer through a window pane:

• Let the temperature difference across the window pane be, $\Delta T$ this temperature difference is the driving cause for the flow of heat, likewise we have "V" the potential difference as the driving cause for the flow of charges.

• Let $\dot{Q}$ be the rate of heat transfer, likewise we have "I" as the rate of flow of charge.

• The heat transfer rate is inversely propotional to $\frac{dx}{k.A}$, where:

       dx is the thickness of the material across which the heat travels,

       k is the thermal conductivity of the material

       A is the area of the pane normal to the heat flow.

       And for current we have a constant factor R called resistance to which         the current is inversely proportional to.

The heat transfer rate is given by the Fourier's Law as:

$\dot Q=k.A\frac{\Delta T}{dx}$

can be further written as

$\dot Q=\frac{\Delta T}{(\frac{dx}{k.A} )}$...............................(1)

The rate of flow of charge i.e. current is given by the Ohm's Law as:

$V=R.I$

$\Rightarrow I=\frac{V}{R}$..............................................(2)

From equations (1) & (2)

$\dot Q\equiv I$

$R\equiv \frac{dx}{k.A}$

$\Delta T\equiv V$