How much heat energy is required to convert 93.4 g of solid ethanol at − 114.5 ° c to gasesous ethanol at 149.8 ° c ? the molar heat of fusion of ethanol is 4.60 kj/mol , and its molar heat of vaporization is 38.56 kj/mol . ethanol has a normal melting point of − 114.5 ° c and a normal boiling point of 78.4 ° c . the specific heat capacity of liquid ethanol is 2.45 j / g ⋅ ° c , and that of gaseous ethanol is 1.43 j / g ⋅ ° c .

Q' = 140.859 kJ

Explanation:

Given that, 93.4 g of solid ethanol at − 114.5 °C is converted to gasesous ethanol at 149.8 ° C.

The molar heat of fusion of ethanol is, ΔH(f) = 4.60 kJ/mol , and its molar heat of vaporization is ΔH(v) = 38.56 kJ/mol .

And also Ethanol has a normal melting point of − 114.5 ° C and a normal boiling point of 78.4 ° C .

The specific heat capacity of liquid ethanol is S(l) = 2.45 J / g ⋅°C , and that of gaseous ethanol is S(g) = 1.43 J / g ⋅°C .

Lets solve this step wise ;

Given 93.4 g of ethanol is taken, but 1 mole of ethanol consists of 46.06 g

⇒ moles of ethanol given = = 2.02 moles

step 1: solid ethanol to liquid ethanol at melting point of − 114.5 ° C

⇒ 1 mole requires ΔH(f) = 4.60 kJ/mol of heat

⇒ heat required = 4.60 × 2.02 = 9.292 kJ.

step 2: liquid ethanol at -114 °C to liquid ethanol at 78.4 °C

Q = m×S×ΔT ; Q = heat required

                       m = mass of the substance

                       S = specific heat of the substance

                       ΔT = change in temperature

Here S = S(l);

⇒ Q = 93.4×2.45×(78.4-(-114.5))

       = 44.141 kJ

step 3: liquid ethanol at 78.4°C to gaseous ethanol at 78.4°C

1 mol of liquid ethanol requires ΔH(v) = 38.56 kJ/mol of heat

⇒ required heat = 38.56×2.02 = 77.89 kJ

step 4: gaseous ethanol at 78.4 °C to gaseous ethanol at 149.8 °C

Q = m×S×ΔT

Here, S = S(g)

Q = 93.4×1.43×(149.8-78.4)

   = 9.536 kJ

⇒ Total heat required = 9.292 + 44.141 + 77.89 + 9.536

                                     = 140.859 kJ

⇒ Q' = 140.859 kJ

The heat loss per unit length is  

Explanation:

From the question we are told that

     The outer diameter of the pipe is

     The thickness is    

      The temperature  of water is    

      The outside air temperature is

        The water side heat transfer coefficient is

       The  heat transfer coefficient is  

The heat lost per unit length is mathematically represented as

Substituting values