The burner on an electric stove has a power output of 2.0 kW. A 650 g stainless steel tea kettle is filled with 20∘C water and placed on the

Question

The burner on an electric stove has a power output of 2.0 kW. A 650 g stainless steel tea kettle is filled with 20∘C water and placed on the already hot burner. If it takes 2.70 min for the water to reach a boil, what volume of water, in cm^3, was in the kettle?

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Tryphena 3 years 2021-07-27T14:33:27+00:00 1 Answers 11 views 0

Answers ( )

  1. Gerda
    0
    2021-07-27T14:34:58+00:00
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    Answer:

    The  value is    V = 900 \ cm^3

    Explanation:

    From the question we are told that

      The  power output is  P_{out} = 2.0 \ kW = 2.0 *10^{3} \ W

       The  mass of the steel is  m= 650 \ g = \frac{650}{1000} = 0.650 \ kg

        The  temperature of the water is T = 20^o C

         The  time take is  t = 2.70 \ minutes = 2.70 *60 = 162 \ s

       

    Generally the quantity of heat energy given out by the  electric stove is mathematically represented as

           Q = P * t

    =>     Q = 2.0 *10^{3} * 162

    =>     Q = 324000 \ J

    This energy can also be mathematically represented as

        Q = \Delta T * m c_s * + m_w * c_w * \Delta T

    Here  c_s is the specific heat of stainless steel with value  c_s = 450\ J/C/kg

     tex]c_s[/tex] is the specific heat of water  with value  c_s = 4180\ J/C/kg

      m_w is the mass of water which is mathematically represented as

          m_w = \rho_w * V

    =>   m_w = 1000 * V

    So

       324000 = (100 -20 ) * 0.650 * 450 * + 1000V * 4180 * (100-20)

        V = 0.0008989 \ m^3

    converting to cm^3

          V = 900 \ cm^3

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