The specific heat of liquid ethanol, C2H5OH(l) is 2.46j/g degree celsius and the heat of vaporization is 39.3 kj/mol the boiling point of ethanol is 78.3 c what amount of etnthalpy is required to heat 50 g of liquid ethanol form 23 c to ethanol vapor a 78.3 c?

The heat required for the given process is 49.52 kJ

Explanation:

The processes involved in the given problem are:  

Pressure is taken as constant.

To calculate the amount of heat absorbed at different temperature, we use the equation:

where,

q = amount of heat absorbed = ?

= specific heat capacity of liquid ethanol = 2.46 J/g°C

m = mass of ethanol = 50 g

= final temperature  = 78.3°C

= initial temperature  = 23°C

Putting values in above equation, we get:

To calculate the amount of heat released at same temperature, we use the equation:

where,

q = amount of heat absorbed = ?

m = mass of ethanol = 50 g

= latent heat of vaporization  = 39.3 kJ/mol

Converting the latent heat of vaporization in J/g, we use the conversion factor:

Molar mass of ethanol = 46 g/mol

1 kJ = 1000 J

So,

Putting values in above equation, we get:

Total heat absorbed =

Total heat absorbed =

Hence, the heat required for the given process is 49.52 kJ

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