An iron wire has a cross-sectional area of 5.00 x 10-6 m2. Carry out steps (a) through (e) to compute the drift speed of the conduction electrons in the wire. (a) How many kilograms are there in 1 mole of iron? (b) Starting with the density of iron and the result of part (a), compute the molar density of iron (the number of moles of iron per cubic meter). (c) Calculate the number density of iron atoms using Avogadro’s number. (d) Obtain the number density of conduction electrons given that there are two conduction electrons per iron atom. (e) If the wire carries a current of 30.0 A, calculate the drift speed of conduction electrons.

The answers at the back of the book are: (a) 55.85 x 10-3 kg/mol, (b) 1.41 x 105 mol/m3, (c) 8.49 x 1028 iron atoms/m3, (d) 1.70 x 1029 conduction electrons/m3, (e) 2.21 x 10-4 m/s.

Explanation:

Given:

Area, A = 5 × 10^-6 m^2

Current, I = 30 A

U = (m × σ × V)/ρ × e × f × l)

= I/(n × A × Q)

Where,

u is again the drift velocity of the electrons, in m⋅s−1

m is the molecular mass of the metal, in kg

σ is the electric conductivity of the medium at the temperature considered, in S/m.

ΔV is the voltage applied across the conductor, in V

ρ is the density (mass per unit volume) of the conductor, in kg⋅m−3

e is the elementary charge, in C

f is the number of free electrons per atom

ℓ is the length of the conductor, in m

A.

Molar mass of iron = 56 g/mol

Converting g to kg,

56 g/mol × 1 kg/1000 g

= 0.056 kg/mol

= approx. 0.056 kilograms in 1 mole of iron.

B.

Density, ρ = 7874 kg/m^3

Molar density = density, ρ/molar mass

= 7874/0.056

= 1.406 × 10^5 mol/m^3

C.

Avogadros constant, Na = 6.022 × 10^23 atoms/mol

Density of iron atoms = avogadros constant × molar density

= 6.022 × 10^23 × 1.406 × 10^5

= 8.478 × 10^28 atoms/m^3

D.

Fe --> Fe2+ + 2e-

density of conduction electrons = 2 conduction electrons/1 atom of iron

= 2 × 8.478 × 10^28

= 1.69 × 10^29 conduction electrons/m^3

E.

Q = 1.602 × 10^-19 C

Using the equation above,

V = 30/(1.69 × 10^29 × 1.602 × 10^-19 × 5 × 10^-6)

= 2.216 × 10^-4 m/s.

c

explanation:

graph slopes whilst gas cools, then it is flat whilst conensation to liquid happens, the coolong liquid then graphs as a slope when it solidifies graph is flat

latent heat