Question

Ryan places 0.150 kg of boiling water in a thermos bottle. How many kgs of ice at –12.0 °C must Ryan add to the thermos so that the equilibrium temperature of the water is 75?a. 0.0436 kgb. 0.0713 kgc. 0.0233 kgd. 0.0265 kge. 0.625 kg

Answer 1

Answer:

The  value  is   $m_i = 0.0234 \ kg$

Explanation:

Generally from the calorimetry principle we have that

      $Heat \ loss\ by\ ice = heat\ gained\ by\ water$

So here heat gained water is mathematically represented as i.e

      $Q_w = m_w * c_w * (T_w - T )$

substituting  0.150 kg for $m_w$ , 4200 J/kg.°C for  $c_w$ , 100°C for   $T_w$ and  75°C for  T

We have  

       $Q_w = 0.150 * 4200 * (100 - 75 )$

         $Q_w =15750 \ J$

The Heat loss by the ice is mathematically represented as

      $Q_i = Q_1 + Q_2 + Q_3$

Here     $Q_1$ is the energy to move the ice to its melting point which is evaluated as  

        $Q_1 = m_i * c_i * ( T_o -T_i)$

Here  $m_i$ is the mass of  ice

       $c_i$ is the specific heat of ice with value  $2.05 * 10^3 J/kg.°C$

          $T_o$ temperature of ice at melting point with value 0°C

           $T_i$ is the temperature of ice with value  -12°C

$Q_2$ is the energy to move the ice from its  its melting point to liquid which is evaluated as  

     $Q_2 = m_i * L$

        Here  L  is the Latent heat of melting of ice with value    $334 * 10^3 J/kg$

$Q_3$ is the energy to move the ice from  liquid  to  the equilibrium temperature  which is evaluated as        

       $Q_1 = m_i * c_w * ( T -T_o)$

So  

     $Q_i = m_i [ c_i * ( T_o -T_i) + L + c_w * ( T -T_o) ]$

=>   $Q_i = m_i [ 2.05 * 10^3 * ( 0 -(-12)) + 334 * 10^3 + 4200 * ( 75 - 0) ]$

From

 $Heat \ loss\ by\ ice = heat\ gained\ by\ water$

We have that

  $m_i * 673600 =15750$

=>     $m_i = \frac{15750}{673600}$

=>     $m_i = 0.0234 \ kg$