Heat can be transferred from one place to another by three different methods: conduction, convection and radiation. Conduction usually occurs in solids convection in liquids and gases. No medium is required for radiation.

Thermal conduction if A is area of cross-section of a conductor I its length K thermal conductivity** T**_{1} and **T**_{2}temperatures at two ends, ten rate of transfer of heat **dQ / dt **or thermal current is given by

**dQ / dt = KA(T _{2} – T_{1}) /1 = – KA dT / dx**

Temperature gradient **dT / dx** is negative in the direction of heat flow.

Comparing it with ohm’s law in electricity **I = V/R**

**I thermal = dQ / dt Vthermal = T _{1} – T_{2} and Rthermal = 1 / KA**

Laws of resistance in series and parallel as in electricity are valid in thermal resistance s also, it is believed that current carriers (free electrons) and heat carriers are same because all electrical conductors are also thermal conductors. In general metals are better thermal conductors than liquids and gases as metals have large number of free electrons.

Thermometric conductivity** (D)** it is the ratio of thermal conductivity to thermal capacity per unit volume. Thus thermometric conductivity or diffusivity is

**D= K/pC where K —–> thermal conductivity **

**P —-> density**

**C —–> specific heat Cv**

Thermal conductivity **K** of gases **K = 1 / 3 V av λp c**_{V}

**= Dp Cv where D = 1 / 3 v av λ **is diffusion coefficient and

**λ = 1/2 πd ^{2}n** is mean free path

**{d—> **effective diameter of a molecule

**{n –>** number of molecules / volume

Wiedemann-franz law Tiedemann and Franz have shown that at a given temperature **T**, the ratio of thermal conductivity **K**) to electrical conductivity **(σ)** is constant

That is,

**K/σT = constant **

**Ingen’s –housz experiment: **Ingen Housz showed that if the number of identical rods of different metals are coated with wax an one of their ends is put in baling water, then in steady state the square of length of the bar over which wax melts is directly proportional to the thermal conductivity of the metal. That is,

**K/L ^{2} = constant **

Thermal resistances in series

**R = R _{1} + R**

_{2}

**I _{1} + I_{2} / K_{eff }A_{eff }+ I_{1} / K_{1} A_{1} + I_{2} / K_{2} A_{2}**

If A

If A

_{1}= A_{2 }and I_{1}= I_{1}

**Then 2/K _{eff} = 1/K_{1} + 1/K**

_{2}

Thermal resistances in parallel

1**/R = 1/R _{1 }+ 1/R_{2}**

K eff A_{eff}/1 = K_{1} A_{1} / 1 + K_{2} A_{2} / 1

Or K_{eff} A_{eff} = K_{1} A_{1} + K_{2} A_{2 }

**And K _{eff} = K_{1} + k_{2}/2**

If the area of cross-section is equal then

**A _{eff} = A_{1} + A_{2} + 2A**

Growth of ice in a pond

**dQ _{1 }= KA 0 – ( – θ) / y dy = dQ_{2 }= mL = PAdyL**

**Or dy / dt = Kθ / pL x 1 / y or 1 = pL y ^{2 }/ Kθ^{2}**

The ratio of times for thicknesses

**0 to y : y to 2y : 2y to 2y : : 1 : 3 : 5**

In a shell of radius **r _{1 }and r_{2}**

**dQ / dt = K 4πr _{1} r_{2} / (r_{2} – r_{2}) (θ_{1} – θ_{2})**

Thickness of the shell **(r _{2} – r_{1}) = K4πr^{2}θ / dQ/ dt**

Where **r _{1} = r_{2} = r and θ_{1} – θ_{2 }= θ**

In a cylinder of length** I, radii r _{1} and r**

_{2}

**dQ / dt = 2πK I ( θ _{1} –n θ_{2})**

**log e r _{2} / r**

_{1}

Expertsmind.com offers physics homework help, physics assignment help, online tutoring, heat transfer problem’s solution and projects assistance with best online support from online qualified and experienced physics tutor’s help.