Question

16. A kettle used at a building site is filled with 25 liters of water at 16 °C. The water is heated until it reaches 90 °C. Calculate the amount of heat required in kJ.

Answer 1

Answer:

The Heat  is  $Q = 7736.7 \ KJ$

Explanation:

Generally from the question we are told that

  The volume of the water is  $V = 25 L = \frac{25}{1000} = 0.025 \ m^3$

   The initial temperature is $T_i = 16^oC = 16 + 273 = 289 \ K$

    The final temperature is $T_f = 90^o C = 90 + 273 = 363 \ K$

Generally the heat required is mathematically represented as

    $Q = m * c_w * (T_f - T_i)$

Here $c_w$  is the specific heat capacity of  water with value  $c_w = 4182 J/(K kg)$

        m  is the mass of the water which is mathematically represented as

         $m = \rho * V$

Here $\rho$ is the density of the water which has a value  $\rho = 1000 kg/m^3$

So

      $m = 0.025 * 1000 = 25 kg$

So

     $Q = 25 * 4182 * (363 - 289)$

=>   $Q = 7736700 \ J$

Converting to KJ

      $Q = \frac{ 7736700}{1000}$

=>    $Q = 7736.7 \ KJ$