Physics, 19.03.2020 03:56 paatnguyyen

The diffusion coefficients for species A in metal B are given at two temperatures: T (°C) D (m2/s) 1030 6.11 × 10-17 1250 6.02× 10-16 (a) Determine the value of the activation energy Qd (in J/mol). J/mol (b) Determine the value of D0. m2/s (Use scientific notation.) (c) What is the magnitude of D at 1160°C? m2/s (Use scientific notation.)

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Answer from: plug30

(a). The value of the activation energy is 171565.80 J/mol.

(b). The value of D₀ is $1.652\times10^{10^{-8}}\ m^2/s$

(c). The magnitude of D at 1160°C is $5.067\times10^{-8}\ m^2/s$

Explanation:

Given that,

Temperatures,

$T_{1}=1030^{\circ}C$

$T_{2}=1250^{\circ}C$

Diffusion coefficients are

$D_{1}=6.11\times10^{-17}$

$D_{2}=6.02\times10^{-16}$

(a). We need to calculate the value of the activation energy

Using relation between diffusion coefficient and temper

$ln(D_{1})=ln(D_{0})-(\dfrac{Q_{d}}{RT_{1}})$ ....(I)

$ln(D_{2})=ln(D_{0})-(\dfrac{Q_{d}}{RT_{2}})$ .....(II)

Divided equation (II) by (I)

$ln(\dfrac{D_{2}}{D_{1}})=\dfrac{Q_{d}}{R}\times(\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}})$

Put the value into the formula

$ln(\dfrac{6.02\times10^{-16}}{6.11\times10^{-17}})=\dfrac{Q_{d}}{8.314}\times(\dfrac{1}{1030+273}-\dfrac{1}{1250+273})$

$Q_{d}=\dfrac{2.2877\times8.314\times1984469}{220}$

$Q_{d}=171565.80\ J/mol$

(b). We need to calculate the value of D₀

Using relation between diffusion coefficient and temperature

$ln(D_{0})=ln(D)+(\dfrac{Q_{d}}{RT_{1}})$

Put the value of in to the formula

$ln(D_{0})=ln(6.11\times10^{-17})+(\dfrac{171565.80}{8.314\times1030+273})$

$ln(D_{0})=-17.9182$

$D_{0}=e^{-17.9182}$

$D_{0}=1.652\times10^{10^{-8}}\ m^2/s$

Now again put the value in the equation (II)

$ln(D)=ln(1.652\times10^{10^{-8}})-(\dfrac{171565.80}{8.314\times1160+273})$

$ln(D_{0})=-16.79776$

$D_{0}=e^{-16.79776}$

$D_{0}=5.067\times10^{-8}\ m^2/s$

Hence, (a). The value of the activation energy is 171565.80 J/mol.

(b). The value of D₀ is $1.652\times10^{10^{-8}}\ m^2/s$

(c). The magnitude of D at 1160°C is $5.067\times10^{-8}\ m^2/s$

Answer from: Quest

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).

Answer from: Quest

change tension

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