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## Conduction of Heat, Physics tutorial

Conduction of Heat Energy:Let a long silver spoon dipped inside the hot soup. After a short time, other end of the spoon becomes hot. Transfer of heat energy through material by conduction could describe this observation. You would have noted that one factor is obvious in conduction of heat. That is, one end is at the higher temperature than other. In the study of conduction of heat through solids, we would like to discuss what factors affect rate of conduction of heat through solids.

Factors Affecting Conduction:Let the state of material with parallel faces of cross-sectional area A at right angles to direction of the flow of heat. Heat Q will flow from high temperature region Θ

_{2}to the low temperature region Θ_{1}. Which means Θ2 is greater than Θ_{1}. Consider distance between the temperatures regions be LIf heat Q takes time t to be transferred across section of material of length L, experimentally it have illustrated that rate at which quantity Q is transferred with respect to time within this region of length L is

i) Proportional to cross-sectional area A:

I.e. Q/t ∝ A .................Eq.1

ii) Proportional to temperature gradient within the region where temperature gradient is stated as

I.e. Q/t ∝ (Θ

_{2}-Θ_{1})/L.................Eq.2Combining two ideas Eq.1 and Eq.2 by saying that the rate of transfer of heat is proportional to the area and temperature gradient, provide:

I.e. Q/t ∝ A X (Θ

_{2}-Θ_{1})/LTherefore Q/t = kA (Θ

_{2}-Θ_{1})/LWhere, k is a constant of proportionality. k is the factor depending on the material.

Thermal Conductivity of a Material:We have been able to state rate at which heat is transferred from the point A1 to point A2 as:

Q/t = kA (Θ

_{2}-Θ_{1)}/LK is hence numerically equivalent to heat transferred per second per unit area of cross-section when unit temperature gradient is set up normal to area.

k = (Rate X heat transferred)/(Area X temperature gradient)

k = [Q/t/(A(Θ

_{2}-Θ_{1})L)]k is a constant of proportionality known as the thermal conductivity for the material for Eq. (7.3) k is written as,

k = (Q/t)(Area x temperature gradient)

(j/s)/(m

^{2}x(^{o}c))= js

^{-1}m^{-1o}C^{-1}= Wm

^{-1o}C^{-1}On in terms of the absolute scale, unit of k can be expressed as

k = Wm

^{-1}K^{-1}Copper, which is a very good conductor of heat, has its value k = 400 Wm

^{-1}K^{-1}and that of air = 0.02 Wm^{-1}K^{-1}One can state thermal conductivity of the material: Thermal conductivity of the material is rate of transfer of heat per unit area per unit temperature gradient through the face of material with the face perpendicular to the direction of the transfer of heat given steady state is maintained.

Lagged and Unlagged Bars:Heat may be transferred under two situations:

Let the conduction of heat through:

Measuring the Thermal Conductivity of a Good Conductor:Searle's bar method is the experimental process to determine thermal conductivity of material. The bar of material is being heated by steam on one side and other side cooled down by water whereas length of the bar is thermally insulated. Then heat ΔQ propagating through bar in the time interval of Δt is provided by

(ΔQ/Δt)

_{bar }= -kA(ΔT_{bar}/L)Where

ΔQ is a heat supplied to bar in time Δt

k is a coefficient of thermal conductivity of a bar.

A is a cross-sectional area of a bar,

ΔT

_{bar }is a temperature difference of both ends of the bar'L is a length of bar

Measuring Thermal Conductivity of the Bad Conductor:The Lee's Disc experiment estimates the approximate value for thermal conductivity k of the poor conductor such as glass, cardboard, etc. Procedure is to put the disc composed of poor conductor, radius r and thickness x, between the steam chamber and two good conductivity metal discs (of same metal) and permit setup to come to equilibrium, so that heat lost by lower disc to convection is similar as heat flow through the poorly conducting disc. Upper disc temperature T

_{2}and lower disc temperature T_{1}are recorded. Poor conductor is removed and lower metal disc is permitted to heat up to upper disc temperature T_{2}. At last, steam chamber and upper disc are removed and replaced by the disc made of the good insulator. Metal disc is then permitted to cool through T_{1}< T_{2}and toward room temperature T_{0}. Temperature of metal disc is recorded as it cools so a cooling curve can be plotted. Then slope s_{1}=ΔT/Δt of cooling curve is estimated graphically where curve passes through temperature T_{1}.Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)Expand your confidence, grow study skills and improve your grades.

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