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 conve

Question

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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Kim Chi 3 years 2021-09-03T08:58:27+00:00 1 Answers 14 views 0

Answers ( )

  1. thuthao
    0
    2021-09-03T09:00:01+00:00
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    Answer:

    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

     So

          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

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