A copper transmission cable 180 km long and 11.0 cm in diameter carries a current of 135 A. Required:
a. What is the potential drop across the cable?
b. How much electrical energy is dissipated as thermal energy every hour?

a) 43.98 V

b) E = 21.37 MJ

Explanation:

Parameters given:

Length of cable = 180 km = 180000 m

Diameter of cable = 11 cm = 0.11 m

Radius = 0.11 / 2 = 0.055 m

Current, I = 135 A

a) To find the potential drop, we have to find the voltage across the wire:

V = IR

=> V = IρL / A

where R = resistance

L = length of cable

A = cross-sectional area

ρ = resistivity of the copper wire = 1.72 * 10^(-8) Ωm

Therefore:

V = (135 * 1.72 * 10^(-8) * 180000) / (π * 0.055^2)

V = 43.98 V

The potential drop across the cable is 43.98 V

b) Electrical energy is given as:

E = IVt

where t = time taken = 1 hour = 3600 s

Therefore, the energy dissipated per hour is:

E = 135 * 43.98 * 3600

E = 21.37 MJ (mega joules, 10^6)

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

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