Respuesta :
Answer: 137.76 A
Explanation: Heat is generated in a material when current is passed through the conducting material, this process is known as Joule heating or Ohmic heating or resistive heating.
Joule's first law, also known as Joule-Lenz law states that the heating power generated by an electrical conductor is directly proportional to the resistance of the conductor and square of the current flowing through it.
Mathematically, we have:
[tex]Q= i^{2} R t[/tex]...............................(1)
where:
- Q= heat generated
- i= current flowing through the conductor
- R= resistance of the conductor
- t= time for which the current flows in the conductor
GIven data:
- heat required for the vapourization of water into steam, L 2256 [tex]J.kg^{-1}[/tex]
- heat of vaporization per kg of water, [tex]c=2256 kJ.kg^{-1}[/tex]
- density of water, [tex]d= 1000 kg.m^{-3}[/tex]
- time for which the current flows= 2.14 ms = [tex]2.14\times 10^{-3}[/tex] s
- resistivity of water, [tex]\rho= 176 \Omega .m[/tex]
- lenght of water along which the current flows, [tex]l=13.2 cm[/tex]
- vertical cross-section area which is normal to the direction of current flow, [tex]A=17.8 \times 10^{-5} m^{2}[/tex]
To find:
- The current required to vaporize the given water, i = ?
Firstly, we calculate the amount of energy required by the given amount of water to vaporize.
We have the density and heat of vaporization per kg of water.
volume of water= [tex]13.2\times 10^{-2}m \times 17.8 \times 10^{-5} m^{2}[/tex]
[tex]Volume= 2.3496 \times 10^{-5} m^{2}[/tex]
Now, Mass of water[tex]= density \times volume\\=1000 \times 2.3496 \times 10^{-5}[/tex]
[tex]Mass= 2.3496 \times 10^{-2} kg[/tex]
∴Total amount of heat energy required,
[tex]Q= m \times L\\Q= 2.3496 \times 10^{-2}\times 2256\\Q= 53.007 J[/tex]
we know, resistance [tex]R= \frac{\rho.l}{A}[/tex]
Putting the given values
[tex]R=\frac{176\times 0.132}{17.8\times 10^{-5} } \\\\R= 130516.85 \Omega[/tex]
Using eq. (1)
[tex]Q= i^{2} R t[/tex]
[tex]53.007 = i^{2} \times 130516.85 \times 2.14\times 10^{-3}\\ \\i^{2} = \frac{53.007}{130516.85 \times 2.14\times 10^{-3}} \\\\i^{2} =0.189781 \\i=0.4356 A[/tex]
