Question

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

Answer 1

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.

Step-by-step explanation:

The process involved in this problem are :

`(1):NH_3(l)(-50^oC)\rightarrow NH_3(l)(-33.4^oC)\\\\(2):NH_3(l)(-33.4^oC)\rightarrow NH_3(g)(-33.4^oC)`

`(3):NH_3(g)(-33.4^oC)\rightarrow NH_3(g)(-0.0^oC)`

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

For process 1 :

`Q_1=m* c_(1)* (T_(final)-T_(initial))`

where,

`Q_1` = amount of heat absorbed = ?

m = mass of ammonia = 13000 g

`c_1` = specific heat of liquid ammonia =
`4.7J/g^oC`

`T_1` = initial temperature =
`-50.0^oC`

`T_2` = final temperature =
`-33.4^oC`

Now put all the given values in
`Q_3`, we get:

`Q_1=13000 g* 4.7 J/g^oC* ((-33.4)-(-50.0))^oC`

`Q_1=1,014,260 J=1.1014.260 kJ`

For process 2 :

`Q_2=n* \Delta H_(fusion)`

where,

`Q_2` = amount of heat absorbed = ?

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

n = Moles of ammonia =
`(13000 g)/(17 g/mol)=764.71 mol`

`\Delta H_(vap)` = enthalpy change for vaporization=23.5 kJ/mol

Now put all the given values in
`Q_1`, we get:

`Q_2=764.71 mol* 23.5 kJ/mol=17,970.6 kJ`

For process 3 :

`Q_3=m* c_(p,l)* (T_(final)-T_(initial))`

where,

`Q_3` = amount of heat absorbed = ?

m = mass of ammonia = 13000 g

`c_2` = specific heat of gaseous ammonia =
`2.2J/g^oC`

`T_1` = initial temperature =
`-33.4^oC`

`T_2` = final temperature =
`0.0^oC`

Now put all the given values in
`Q_3`, we get:

`Q_3=13000 g* 2.2J/g^oC* (0.0-(-33.4))^oC`

`Q_3=955,240 J=955.240 kJ`

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=Q_1+Q_2+Q_3=17,970.6 kJ+17,970.6 kJ+955.240 kJ`

Q = 36,896.44 kJ