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3 years ago

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The mass fraction of eutectoid cementite in a hypoeutectoid iron-carbon alloy (just below its eutectoid temperature) is 0.109. O

n the basis of this information, determine the composition of the alloy. If it is not possible to determine the composition from the information provided, enter 0.

1 answer:

Zinaida [17]3 years ago

3 0

Answer:

The composition of an alloy is 0.75%wt

Explanation:

Let alloy is a hypoeutectoid alloy.

So, we can apply lever rule which is shown below.

$W_{a}=$ $\frac{C_{0} -C_{a} }{C_{b}-C_{a} }$

We know that $C_{a}=0.022$ and $C_{b}=6.7$

Given that $W_{a}=0.109$ , we have to find $C_{0}$

Thus,

$0.109=\frac{C_{0}-0.022 }{6.7-0.022}$

Hence

$C_{0}=0.75$ %wt

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Convert 850 nm wavelength into frequency, eV, wavenumber, joules and ergs.

Sholpan [36]

Answer:

Frequency = $3.5294\times 10^{14}s^{-1}$

Wavenumber = $1.1765\times 10^6m^{-1}$

Energy = $2.3365\times 10^{-19}J$

Energy = 1.4579 eV

Energy = $2.3365\times 10^{-12}erg$

Explanation:

As we are given the wavelength = 850 nm

conversion used : $(1nm=10^{-9}m)$

So, wavelength is $850\times 10^{-9}m$

The relation between frequency and wavelength is shown below as:

$Frequency=\frac{c}{Wavelength}$

Where, c is the speed of light having value = $3\times 10^8m/s$

So, Frequency is:

$Frequency=\frac{3\times 10^8m/s}{850\times 10^{-9}m}$

$Frequency=3.5294\times 10^{14}s^{-1}$

Wavenumber is the reciprocal of wavelength.

So,

$Wavenumber=\frac{1}{Wavelength}=\frac{1}{850\times 10^{-9}m}$

$Wavenumber=1.1765\times 10^6m^{-1}$

Also,

$Energy=h\times frequency$

where, h is Plank's constant having value as $6.62\times 10^{-34}J.s$

So,

$Energy=(6.62\times 10^{-34}J.s)\times (3.5294\times 10^{14}s^{-1})$

$Energy=2.3365\times 10^{-19}J$

Also,

$1J=6.24\times 10^{18}eV$

So,

$Energy=(2.3365\times 10^{-19})\times (6.24\times 10^{18}eV)$

$Energy=1.4579eV$

Also,

$1J=10^7erg$

So,

$Energy=(2.3365\times 10^{-19})\times 10^7erg$

$Energy=2.3365\times 10^{-12}erg$

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A rigid, well-insulated tank of volume 0.9 m is initially evacuated. At time t = 0, air from the surroundings at 1 bar, 27°C beg

Eva8 [605]

Answer:

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Explanation:

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From first law of thermodynamics for unsteady flow process

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Given that tank is insulated so $\dot{Q}=0$ and no mass is leaving so

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Mass conservation $m_2-m_1=m_e-m_i$

$m_1,m_2$ is the initial and final mass in the system respectively.

Initially tank is evacuated so $m_1=0$

We know that for air $u=C_vT ,h=C_p T$ , $P_2v_2=m_2RT_2$

$m_2=0.42 kg$

So now putting values

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$\dot{w}$ = -0.303 KW

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