Question

Ethanol (C2H5OH) melts at –114 °C and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kJ/mol, and its enthalpy of vaporization is 38.56 kJ/mol. The specific heats of solid and liquid ethanol are 0.97 J/g-K and 2.3 J/g-K, respectively. The average specific heat of gaseous ethanol is about 1.80 J/g-K. a. How much heat is required to convert 35.0 g of ethanol at 27 °C to the vapor phase at 120 °C? b. How much heat is required to convert the same amount of ethanol at –120 °C to the vapor phase at 120 °C?

Answer 1

Answer:

First question

       $Q = 36826 \ J$

Second  question  

       $Q = 52299.7 \ J$

Explanation:

From the question we are told that

     The melting point of Ethanol is  $T_m = -114 ^oC$

      The boiling point of Ethanol is  $T_b = 78^ oC$

       The enthalpy of fusion of Ethanol is $F = 5.02 \ kJ / mol = 5.02 *10^{3}\ kJ / mol$

        The enthalpy of vaporization  of Ethanol is $L = 38.56 \ kJ / mol = 38.56 *10^{3} \ J / mol$

         The specific heat of solid Ethanol is  $c_e = 0.97 \ J/ g \cdot K$

          The specific heat of liquid  Ethanol is $c_l = 2.3 \ J / g \cdot K$

           The mass of the Ethanol given is  $m = 35.0 \ g$

Considering the first question

           The initial  temperature is $T_i = 27^oC$

             The final  temperature is  $T_f = 120^oC$

Generally the heat required too raise the Ethanol to its boiling point is mathematically represented as

       $Q_1 = m * c_l * (T_b - T_i)$

=>      $Q_1 = 35.0 * 2.3 * ( 78 - 27)$

=>      $Q_1 =4106 \ J$

Genially the number of moles of Ethanol given is mathematically represented as

         $n = \frac{m}{Z}$

Here Z  is the molar mass of Ethanol  with value  $Z = 46 g/mol$

So

         $n = \frac{35}{46 }$

=>      $n = 0.7609 \ mol$

Generally the heat of vaporization of the Ethanol is mathematically represented as

         $Q_2 = n * L$

=>        $Q_2 =0.7809 * 38.56 * 10^{3}$

=>        $Q_2 =29339 \ J$

Generally the heat required too raise the Ethanol from  its boiling point to  $T_f$  is  mathematically represented as

       $Q_3 = m * c_l * (T_f - T_b)$

=>     $Q_3 = 35 * 2.3 * (120 - 78 )$

=>     $Q_3 = 3381 \ J$

Generally the total heat required is  

     $Q = Q_1 + Q_2 + Q_3$

=>   $Q = 4106 + 29339 + 3381$

=>   $Q = 36826 \ J$

Considering the second question

           The initial  temperature is $T_i = -120^oC$

             The final  temperature is  $T_f = 120^oC$

Generally the heat required too raise the Ethanol to its melting  point is mathematically represented as

       $Q_1 = m * c_e * (T_m - T_i)$

=>      $Q_1 = 35.0 * 0.97 * ( -114 - (- 120) )$

=>      $Q_1 = 203.7 \ J$

Generally the heat of fusion  of the Ethanol is mathematically represented as

                 $Q_2 = n * F$

=>        $Q_2 =0.7809 * 5.02 *10^{3}$

=>        $Q_2 =3920 \ J$

Generally the heat required too raise the Ethanol to its boiling point is mathematically represented as

       $Q_3 = m * c_l * (T_b - T_m)$

=>      $Q_3 = 35.0 * 2.3 * ( 78 - (- 114) )$

=>      $Q_3 =15456 \ J$

Generally the heat of vaporization of the Ethanol is mathematically represented as

         $Q_4 = n * L$

=>        $Q_4 =0.7809 * 38.56 * 10^{3}$

=>        $Q_4 =29339 \ J$

Generally the heat required too raise the Ethanol from  its boiling point to  $T_f$  is  mathematically represented as

       $Q_5 = m * c_l * (T_f - T_b)$

=>     $Q_5 = 35 * 2.3 * (120 - 78 )$

=>     $Q_5 = 3381 \ J$

Generally the total heat required is  

     $Q = Q_1 + Q_2 + Q_3+Q_4 + Q_5$

=>   $Q = 203.7 + 3920 + 15456 +29339+3381$

=>   $Q = 52299.7 \ J$