Physics, 05.06.2020 21:59, mova1o

The heating element in an electric kettle is rated as 2.0 kW. If the water in the kettle is at 100.0 °C, what volume of water will be converted into steam in one minute? The specific latent heat of vaporization of the water is 2,257,000 J/kg and the 3 density of water is 1,000 kg/m .

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answered: IntellTanito

The volume is $V =5.32 *10^{-5} \ m^3$

Explanation:

From the question we are told that

The power of the heating element is $P = 2.0 kW = 2.0 *10^3 \ W$

The temperature of the water in the kettle is T _w = 100^oC

The time to convert water to steam is t = 1 minute = 60 sec

The specific latent heat of vaporization is $H_v = \ 2,257,000 J/kg$

The density of water is $\rho_w = 1000\ kg/m^3$

The power of the heating element is mathematically represented as

$P = \frac{E}{t}$

Where E Energy generated by the heating element in term of heat

E = Pt

substituting values

$E = 2.0 *10^{3} * 60$

E = 120000 J

Now

The latent heat of vaporization is mathematically represented as

$H_v = \frac{E}{m}$

Where m is the mass of water converted to steam

$m = \frac{E}{H_v}$

substituting values

$m = \frac{120000}{2257000}$

$m = 0.0532\ kg$

The volume of water converted to steam is mathematically evaluated as

$V = \frac{m }{\rho_w}$

substituting values

$V = \frac{0.0532}{1000}$

$V =5.32 *10^{-5} \ m^3$

answered: Guest

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answered: Guest

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The heating element in an electric kettle is rated as 2.0 kW. If the water in the kettle is at 100.0...