All categories

3 years ago

What is the temperature dependency of the electrical conductivity for metals and semiconductors??

Engineering

1 answer:

geniusboy [140]3 years ago

3 0

Answer and Explanation:

TEMPERATURE DEPENDENCY ON ELECTRICAL CONDUCTIVITY OF METALS : Metals are good conductors of electricity but when we increase the temperature the free electrons of metals collide with each other due to heat.There collision become very fast and so the resistance increases and so the electrical conductivity of metals decreases on increasing temperature.

TEMPERATURE DEPENDENCY ON ELECTRICAL CONDUCTIVITY OF SEMICONDUCTOR : The electrical conductivity of semiconductor is mainly sue to presence of impurities and defects as the temperature increases the impurities and defects also increases so the electrical conductivity of semiconductor increases on increasing temperature.

You might be interested in

You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19

Pepsi [2]

Answer:

To answer this question we assumed that the area units and the thickness units are given in inches.

The number of atoms of lead required is 1.73x10²³.

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

$V = A*t$

Where:

A: is the surface area = 160

t: is the thickness = 0.002

Assuming that the units given above are in inches we proceed to calculate the volume:

$V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}$

Now, using the density we can find the mass:

$m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g$

Finally, with the Avogadros number ( $N_{A}$ ) and with the atomic mass (A) we can find the number of atoms (N):

$N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms$

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

3 0

2 years ago

A single-phase transformer circuit feeds a motor and lighting load of 50 kilowatts. At a power factor of .8, the KVA rating of t

AveGali [126]

The KVA rating of the step down transformer at the given power factor would be 62.5 kVA.

<h3>What is power factor of a transformer?</h3>

Power factor (PF) is the ratio of working power, measured in kilowatts (kW), to apparent power, measured in kilovolt amperes (kVA).

PF = working power / apparent power

PF = kW/kVA

kVA = kW/PF

kVA = 50 kW/0.8

kVA = 62.5 kVA

Thus, the KVA rating of the step down transformer at the given power factor would be 62.5 kVA.

Learn more about power factor here: brainly.com/question/7956945

#SPJ1

3 0

1 year ago

A lake is fed by a polluted stream and a sewage outfall. The stream and sewage wastes have a decay rate coefficient (k) of 0.5/d

joja [24]

Solution :

Given :

k = 0.5 per day

$C_s = 10 \ mg/L \ ; \ \ Q_s= 40 \ m^3/s$

$C_{sw} = 100 \ ppm \ ; \ \ Q_{sw}= 0.5 \ m^3/s$

Volume, V $= 200 \ m^3$

Now, input rate = output rate + KCV ------------- (1)

Input rate $= Q_s C_s + Q_{sw}C_{sw}$

$=(40 \times 10) + (0.5\times 100)$

$= 2 \times 10^5 \ mg/s$

The output rate $= Q_m C_{m}$

= ( 40 + 0.5 ) x C x 1000

$=40.5 \times 10^3 \ C \ mg/s$

Decay rate = KCV

∴ $KCV =\frac{0.5/d \times C \ \times 200 \times 1000}{24 \times 3600}$

= 1.16 C mg/s

Substituting all values in (1)

$2 \times 10^5 = 40.5 \times 10^3 \ C+ 1.16 C$

C = 4.93 mg/L

4 0

2 years ago

In a production facility, 3 cm thick large brass plates (k = 110 W/mC, α = 33.9 × 10-6 m2 /s) that are initially at a uniform

zysi [14]

Answer:

Explanation:

Given.

Thickness of brass plate $t = 3 cm$

Thermal conductivity of brass $k = 110 W/m.°C$

Density of brass $\rho = 8530 kg/m^3$

Specific heat of brass $C_p =380J/kg.°C$

Thermal diffusivity of brass $\alpha = 33.9\times 10^{-6} m^2/s$

Temperature of oven $T_{\infty} = 700°C$

The initial temperature $T_i= 25°C$

Plate remain in the oven $t =10 min$

Heat conduction in the plate is one-dimensional since the plate is large relative to its thickness and there is thermal symmetry about the center plane.

The thermal properties of the plate are constant.

The heat transfer coefficient is constant and uniform over the entire surface.

The Fourier number is > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

The Biot number for this process $Bi = \frac{hL}{k}\\\\Bi=\frac{(80 W/m^2.°C)(0.015 m)}{(110 W/m.°C)}\\=Bi =0.0109$

The constants $\lambda_1$ and $A_1$ corresponding to this Biot are, from 11-2 tables.

The interpolation method used to find the

$\lambda_1=0.1039$ and $A_1=1.0018
$

The Fourier number $\tau=\frac{\alpha t}{L^2}\\\\\tau=\frac{(33.9\times 10^{-6} m^2/s)(10 min \times 60 s/min)}{(0.015m)^2}
\\\\\tau=90.4>0.2$

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable.

Then the temperature at the surface of the plates becomes

$\theta(L,t)_{wall}=\frac{T(x,t)-T_{\infty}}{(T_i-T_{\infty})}\\\\\theta(L,t)_{wall}=A_1e^{-\lambda_1^2\tau}\cos(\lambda_1L/L)\\\\\theta(L,t)_{wall}=(1.0018)e^{-(0.1039^2(90.4))}\cos(0.1039)\\\\\theta(L,t)_{wall}=0.378\\\\\frac{T(L,t)-700}{25-700}=0.378\\\\T(L,t)=445°C$