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?

Answer 1

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 $cm^3$

      $V = 900 \ cm^3$