Question

How much water

Answer 1

Answer:

The mass is  $m = 3.75 \ kg$

Explanation:

From the question we are told that

     The initial temperature is  $T_1 = 15^oC$

      The final  temperature is  $T_2 = 100 ^o C \ (boiling point )$

Generally the maximum heat produced by  1  Liter   of  natural gas is  $9000 \ cal$

 So the amount of heat produced by 100 L is  

              $E = 9000 * 100$

=>           $E = 9000 00 \ cal$

Generally given that the efficiency is  $\eta = 0.35$

Then actual heat received by the water is

         $H = 0.35 * E$

=> $H = 0.35 * 9000 00$

=> $H = 315000 \ cal$

Converting to kcal  

=> $H = 315000 \ cal = \frac{315000}{1000} = 315 \ kcal$

Generally the specific heat of water is  

       $c_w = 1 kcal/ kg \cdot ^oC$

Generally the heat received by the water is mathematically represented as

       $H = m * c_w * (T_2 - T_1)$

=>     $315 = m * 1 * ( 100 - 15 )$

=>     $m = 3.75 \ kg$