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ch4aika [34]
3 years ago
11

A thick oak wall (rho = 545 kg/m3 , Cp = 2385 J/kgK, and k = 0.17 W/mK) initially at 25°C is suddenly exposed to combustion prod

ucts at 800°C. Determine the time of exposure necessary for the surface to reach the ignition temperature of 400°C, assuming the convection heat transfer coefficient between the wall and the products to be 20 W/m2 K. At that time, what is the temperature 1 cm below the surface? (Note: use an appropriate equation for the semi-infinite wall case; compare equations 18.20 and 18.21 in the text).

Engineering
1 answer:
Nimfa-mama [501]3 years ago
8 0

Answer:

Explanation:

The detailed calculation and appropriate equation with substitution is as shown in the attached file.

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A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 m
natima [27]

Answer:

a. 0.4544 N

b. 5.112 \times 10^{-5 M}

Explanation:

For computing the normality and molarity of the acid solution first we need to do the following calculations

The balanced reaction

H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O

NaOH\ Mass = Normality \times equivalent\ weight \times\ volume

= 0.3200 \times 40 g \times 21.30 mL \times  1L/1000mL

= 0.27264 g

NaOH\ mass = \frac{mass}{molecular\ weight}

= \frac{0.27264\ g}{40g/mol}

= 0.006816 mol

Now

Moles of H_2SO_4 needed  is

= \frac{0.006816}{2}

= 0.003408 mol

Mass\ of\ H_2SO_4 = moles \times molecular\ weight

= 0.003408\ mol \times 98g/mol

= 0.333984 g

Now based on the above calculation

a. Normality of acid is

= \frac{acid\ mass}{equivalent\ weight \times volume}

= \frac{0.333984 g}{49 \times 0.015}

= 0.4544 N

b. And, the acid solution molarity is

= \frac{moles}{Volume}

= \frac{0.003408 mol}{15\ mL \times  1L/1000\ mL}

= 0.00005112

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We simply applied the above formulas

4 0
3 years ago
After a strong storm, a worker does not realize that a power transmission line has fallen on his car and is electrocuted while o
joja [24]

Answer and Explanation:

<u>The correct answer choice is Contacting Overhead Power Lines.</u>

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Explanation:

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Bezzdna [24]

Answer:

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Explanation:

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____ [38]

Answer:

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Hope this helped you!

Explanation:

5 0
3 years ago
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