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Ne4ueva [31]
3 years ago
5

(a) Aluminum foil used for storing food weighs about 0.3 grams per square inch. How many atoms of aluminum are contained in one

square inch of the foil? (b) Using the densities and atomic weights given in Appendix A, calculate and compare the number of atoms per cubic centimeter in (i) lead and (ii) lithium.

Engineering
1 answer:
atroni [7]3 years ago
3 0

Answer:

note:

solution is attached due to error in mathematical equation. please find the attachment

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Consider a very long, cylindrical fin. The temperature of the fin at the tip and base are 25 °C and 50 °C, respectively. The dia
Mrrafil [7]

Answer:

The fin temperature in °C at a distance of 10 cm from the base = 33.78°C

Explanation:

The following assumptions will be made to solve this problem

- The heat transfer coefficient does not change with the time or distance.

- The temperature of the fins varies just in only one direction.

The temperature of the fin at x = 10 cm = 0.10 m from the base can be calculated from the temperature variation with distance formula for a very long fin.

(T - T∞) = (T₀ - T∞)e⁻ᵐˣ

T = T(x) = temperature at any point along the fin

T∞ = temperature at the tip of the fin = ambient temperature = 25°C

T₀ = temperature at the base of thw fin = 50°C

x = any distance along the length of the fin from the base of the fin = 0.1 m

m = √(hP/KA)

h = Heat transfer coefficient = 123 W/m².K

P = perimeter in contact with the base = πD = π × 0.03 = 0.0943 m

K = thermal conductivity = 150 W/m.K

A = surface area in contact with the base = πD²/4 = π(0.03)²/4 = 0.0007071 m²

m = √(123 × 0.0943)/(150 × 0.0007071)

m = 10.46

mx = 10.46 × 0.1 = 1.046

(T - 25) = (50 - 25) e⁻¹•⁰⁴⁶

T = 25 + 25 e⁻¹•⁰⁴⁶ = 25 + 8.78 = 33.78°C

8 0
3 years ago
How do I draw this from the side?
tangare [24]

Answer:

draw it 3D

Explanation:

because it's a 3D picture

6 0
3 years ago
If copper (which has a melting point of 1085°C) homogeneously nucleates at 849°C, calculate the critical radius given values of
____ [38]

Answer:

The critical radius is -1.30 nm

Explanation:

Temperature for homogenous nucleation of copper, T_{H} = 849^{0} C = 849 + 273 = 1122 K

Melting point of copper, T_{cu} = 1085^{0} C = 1085 + 273 = 1358 K

Latent heat of fusion, H_{f} = -1.77 * 10^{9} J/m^{3}

Surface free energy, \gamma = 0.200 J/m^{2}

Critical radius, r = ?

The formula for the critical radius is given by:

r = \frac{2 \gamma T_{cu} }{H_{f}(T_{cu} - T_{H})  }

r = \frac{2 * 0.2*1358 }{(-1.77 * 10^{9}) (1358 - 1122)  }

r = \frac{543.2 }{(-1.77 * 10^{9}) 236}\\r = -1.30 * 10^{-9} m\\r = -1.30 nm

The critical radius is -1.30 nm

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4 years ago
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3 years ago
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