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 6 months 2021-07-27T14:33:27+00:00 1 Answers 7 views 0

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    2021-07-27T14:34:58+00:00

    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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