Liquid ammonia, , was used as a refrigerant fluid before the discovery of the chlorofluorocarbons and is still widely used today. its normal boiling point is –33.4 °c, and its vaporization enthalpy is 23.5 kj/mol. the gas and liquid have specific heat capacities of 2.2 and 4.7 , respectively. calculate the heat energy transfer required to raise the temperature of 13.0 kg liquid ammonia from –50.0 °c to 0.0 °c. heat energy = kj

The heat energy transfer required to raise the temperature of 13.0 kg liquid ammonia from -50.0 °C to 0.0 °C is 36,896.44 kJ.

Explanation:

The process involved in this problem are :

Now we have to calculate the amount of heat released or absorbed in both processes.

For process 1 :

where,

= amount of heat absorbed = ?

m = mass of ammonia = 13000 g

= specific heat of liquid ammonia  =

= initial temperature =

= final temperature =

Now put all the given values in , we get:

For process 2 :

where,

= amount of heat absorbed = ?

m = mass of solid ammonia = 13.0 Kg  = 13000 g

n = Moles of ammonia =

= enthalpy change for vaporization=23.5 kJ/mol

Now put all the given values in , we get:

For process 3 :

where,

= amount of heat absorbed = ?

m = mass of ammonia = 13000 g

= specific heat of gaseous ammonia  =

= initial temperature =

= final temperature =

Now put all the given values in , we get:

The heat energy transfer required to raise the temperature of 13.0 kg liquid ammonia from -50.0 °C to 0.0 °C  = Q

Q = 36,896.44 kJ

possibly, yes.

explanation:

the amount of sugar you take in is dangerously high, risk of heart disease, high risk of diabetes, and etc.

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.