Heat Transfer Concept Page - 2

Example

Thermal resistance for a given arrangement

Three rods of material x$x$ and three rods of material y$y$ are connected as shown in the figure. All the rods are of identical length and cross section. If the end A is maintained at 60o${}^o$C and the junction E at 10o${}^o$C, find the effective thermal resistance. Given length of each rod=$=$ l$l$, area of cross-section=A$=A$, conductivity of x=K$x=K$ and conductivity of y=2K$y=2K$.Thermal Resistivity is defined as R=LKA$R=\frac{L}{KA}$
Rx=LKA$R_x={\displaystyle \frac{L}{KA}}$
Ry=L2KA$R_y={\displaystyle \frac{L}{2KA}}$
If  we take Rx=2R$R_x=2R$ then, Ry=R$R_y=R$
Since the bridge is balanced i.e  RBCRCE=RBDRDE${\displaystyle \frac{R_{BC}}{R_{CE}}}={\displaystyle \frac{R_{BD}}{R_{DE}}}$
the temp of junctions C and D is equal and the rod CD becomes ineffective as no heat will flow through it.
The thermal resistance of the bridge between junction B and E is
1RBE=12R+2R+1R+R${\displaystyle \frac{1}{R_{BE}}}={\displaystyle \frac{1}{2R+2R}}+{\displaystyle \frac{1}{R+R}}$
∴RBE=43R$\therefore R_{BE}={\displaystyle \frac{4}{3}}R$
The total resistance between A and E will be,
Req=RAB+RBE=R+43R=73R$R_{eq}=R_{AB}+R_{BE}=R+{\displaystyle \frac{4}{3}}R={\displaystyle \frac{7}{3}}R$
Substituting R=L2KA$R={\displaystyle \frac{L}{2KA}}$ we get Req=73×L2KA=7L6KA$R_{eq}={\displaystyle \frac{7}{3}}\times {\displaystyle \frac{L}{2KA}}={\displaystyle \frac{7L}{6KA}}$

Definition

Searle's cone method

As the counter is rotated there is friction between the vessels which bring about a torque τ$\tau$ which is opposed by the torque of the pan and its weights = mgR.
τ=mgR$\tau =mgR$ when m is adjusted so that the inner vessel doesn't move.
Suppose,
mass of water taken = m1$m_1$
mass of the two vessels taken together = m2$m_2$
mass of the pan and its weights = m$m$
initial temperature of water = T1$T_1$
final temperature of water = T2$T_2$
radius of the disc = R
no of revolutions made by the outer vessel = n
specific heat capacity of water = s1$s_1$
specific heat capacity of vessels' materials = s2$s_2$
Then work done by the torque = mgR2πn$mgR2\pi n$
Heat needed to increase the temperature of water=m1s1(T2−T1)$m_1{s}_1(T_2-T_1)$
Heat needed to increase the temperature of the vessels = m2s2(T2−T1)$m_2{s}_2(T_2-T_1)$
Thus mgR2πn=J(m1s1+m2s2)(T2−T1)$mgR2\pi n=J(m_1{s}_1+m_2{s}_2)(T_2-T_1)$
J=mgR2πnJ(m1s1+m2s2)(T2−T1)=4.186Jcal−1$J={\displaystyle \frac{mgR2\pi n}{J(m_1{s}_1+m_2{s}_2)(T_2-T_1)}}=4.186Jca{l}^{-1}$

Definition

Searle's method in calculation of thermal conductivity

Searle's bar method is an experimental procedure to measure thermal conductivity of material. A bar of material is being heated by steam on one side and the other side cooled down by water while the length of the bar is thermally insulated. Then the heat Q propagating through the bar in a time interval of t is given by:
(ΔQΔT)bar=−kAΔTbarL$({\displaystyle \frac{\mathrm{\Delta }Q}{\mathrm{\Delta }T}}{)}_{bar}=-kA{\displaystyle \frac{\mathrm{\Delta }{T}_{bar}}{L}}$
Q is the heat supplied to the bar in time t
k is the coefficient of thermal conductivity of the bar.
A is the cross-sectional area of the bar,
Tbar${}_{bar}$ is the temperature difference of both ends of the bar
L is the length of the bar and
the heat Q absorbed by water in a time interval of t is:
(ΔQΔt)water=CwΔmΔtΔTwater${\left(\frac{\mathrm{\Delta }Q}{\mathrm{\Delta }t}\right)}_{water}=C_w\frac{\mathrm{\Delta }m}{\mathrm{\Delta }t}{\mathrm{\Delta }T}_{water}$
where Cw$C_w$ is the specific heat of water,
m is the mass of water collected during time t,
Twater${}_{water}$ is difference in the temperature of water before and after it has gone through the bar.
Assuming perfect insulation and no energy loss, then
(ΔQΔt)bar=(ΔQΔt)water${\left({\displaystyle \frac{\mathrm{\Delta }Q}{\mathrm{\Delta }t}}\right)}_{bar}={\left({\displaystyle \frac{\mathrm{\Delta }Q}{\mathrm{\Delta }t}}\right)}_{water}$
which leads to 
k=−CwLAΔmΔtΔTwaterΔTbar$k=-C_w{\displaystyle \frac{L}{A}}{\displaystyle \frac{\mathrm{\Delta }m}{\mathrm{\Delta }t}}{\displaystyle \frac{{\mathrm{\Delta }T}_{water}}{{\mathrm{\Delta }T}_{bar}}}$

Definition

Wiedemann Franz law

The WiedemannFranz law is the ratio of the electronic contribution of the thermal conductivity (k$k$) to the electrical conductivity (σ$\sigma$) of a metal, and is proportional to the temperature (T$T$).
kσ=LT${\displaystyle \frac{k}{\sigma }}=LT$
Theoretically, the proportionality constant L$L$, known as the Lorenz number, is equal to
L=kσT=π23(kBe)2=2.44×10−8WΩK−2.${\displaystyle L=\frac{k}{\sigma T}=\frac{{\pi }^2}{3}{\left(\frac{k_B}{e}\right)}^2=2.44\times {10}^{-8}\mathrm{W}\mathrm{\Omega }{\mathrm{K}}^{-2}.}$

Example

Uses of good conductors

The following are some uses of good conductors of heat

1. Cooking utensils are made from good conductors of heat such as copper, brass, steel, aluminium etc.,as the  utensils get heated up quickly and food can be cooked efficiently in shorter time. 
2. Mercury is used in a thermometer as it is a very good conductor of heat such that it acquires the temperature to be measured by the thermometer.
3. Copper being a good conductor of heat is used in electric geyser, to conducts heat from the heated element to the water in the geyser.
4. The bit of a soldering iron is made of copper. As the tip of the hot bit comes into contact with the material  to be soldered, heat is rapidly conducted from its heated parts to its tip. So the temperature of the tip becomes more than the melting point of the soldering material and the soldering material  gets melted soon.

Example

Experimental verification of air as a bad conductor of heat

Take a very hot Bunsen flame. Hold a live matchstick about 1cm away from the Bunsen flame. The matchstick does not get hot enough to burst into flame. This shows that air is a very poor conductor of heat.

Definition

Radiation from Black and White Surfaces

Particular types of clothes are suitable for a particular season.For example, In winters, woolen clothes and dark colors clothes are preferable as it absorbes heat, where as in summer season, light color cotton clothes are preferable as it reflects heat and keep the body cool.

Definition

Woolen Clothes Keeps People Warm in Winter

Wool is bad conductor of heat, it provides insulation from temperature.It traps the heat inside the clothes in order to provide warmth during winter.But wearing woolen cloths in summer would be terrible as the woolen cloths are water repellent and will not absorb the sweat during summer to keep the body dry.

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