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Gennadij [26K]
3 years ago
7

The weatherman states that the current temperature is 95 F and the dew point is 74 F. What is the current relative humidity?

Engineering
1 answer:
spayn [35]3 years ago
3 0

Answer:

Current Relative Humidity is 29.623

Given:

Current Temperature, T_{c} = 95 F = 35^{\circ}C

Dew point temperature, T_{d} = 74 F = 23.34^{\circ}C

Solution:

Now, in order to calculate the Relative Humidity, RH, we use the given formula:

T_{d} =100( \frac{\frac{aT_{c}}{b + T}b + lnRH}{a - \frac{aT_{c}}{b + T} + lnRH})

where

a = 17.625

b = 237.7

Now, using the above formula and given values:

23.34 = (237.7\frac{\frac{17.625\times 35}{ 237.7 + 35} + lnRH}{17.625 - \frac{17.625\times 35}{237.7 + 35} + lnRH})

23.34(17.625 - 2.26 + lnRH) = (237.7\times 2.26 + lnRH)

358.62 + 23.34lnRH) = (537.20 + lnRH)

358.62 + 23.34lnRH) = (537.20 + lnRH)

On solving the above eqn, we get:

RH = 29.623

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Tin atoms are introduced into an FCC copper ,producing an alloy with a lattice parameter of 4.7589×10-8cm and a density of 8.772
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Answer:

atomic percentage = 143 %

Explanation:

Let  x be the number of tin atoms and there are 4 atoms / cell in the FCC structure , then 4 -x  be the number of copper atoms . Therefore, the value of x can be determined by using the density equation as shown below:

\mathbf{density (\rho) = \dfrac{(no \ of \ atoms/cell)(atomic \ mass )}{(lattice \ parameter )^3(6.022*10^{23} atoms/ mol)} }

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\mathbf{8.772 g/cm^3 = \dfrac{(x)(118.69 \ g/mol) +(4-x)(63.54 \ g/mol)}{(4.7589*10^{-8} cm )^3(6.022*10^{23} atoms/ mol)} }

569.32 = 118.69x + 254.16-63.54x

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As there are four atoms per cell in FCC structure for the metal, thus, the atomic percentage of the tin is  calculated as follows :

atomic % = \frac{no \ of \ atoms \ per  \ cell \ in \ tin }{no \ of \ atoms \ per  \ cell \ in \ the \ metal}*100

atomic % = \frac{5.72 \ atoms / cell}{4 \ atoms/ cell} *100

atomic % = 143 %

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