Answer:
The volume is [tex]V =5.32 *10^{-5} \ m^3[/tex]
Explanation:
From the question we are told that
The power of the heating element is [tex]P = 2.0 kW = 2.0 *10^3 \ W[/tex]
The temperature of the water in the kettle is [tex]T _w = 100^oC[/tex]
The time to convert water to steam is t = 1 minute = 60 sec
The specific latent heat of vaporization is [tex]H_v = \ 2,257,000 J/kg[/tex]
The density of water is [tex]\rho_w = 1000\ kg/m^3[/tex]
The power of the heating element is mathematically represented as
[tex]P = \frac{E}{t}[/tex]
Where E Energy generated by the heating element in term of heat
[tex]E = Pt[/tex]
substituting values
[tex]E = 2.0 *10^{3} * 60[/tex]
[tex]E = 120000 J[/tex]
Now
The latent heat of vaporization is mathematically represented as
[tex]H_v = \frac{E}{m}[/tex]
Where m is the mass of water converted to steam
So
[tex]m = \frac{E}{H_v}[/tex]
substituting values
[tex]m = \frac{120000}{2257000}[/tex]
[tex]m = 0.0532\ kg[/tex]
The volume of water converted to steam is mathematically evaluated as
[tex]V = \frac{m }{\rho_w}[/tex]
substituting values
[tex]V = \frac{0.0532}{1000}[/tex]
[tex]V =5.32 *10^{-5} \ m^3[/tex]
