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

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer