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

Answer:

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 :

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

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

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

For process 1 :

$Q_1=m\times c_{1}\times (T_{final}-T_{initial})$

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:

$Q_1=13000 g\times 4.7 J/g^oC\times ((-33.4)-(-50.0))^oC$

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

For process 2 :

$Q_2=n\times \Delta H_{fusion}$

where,

$Q_2$ = amount of heat absorbed = ?

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

n = Moles of ammonia = $\frac{13000 g}{17 g/mol}=764.71 mol$

$\Delta H_{vap}$ = enthalpy change for vaporization=23.5 kJ/mol

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

$Q_2=764.71 mol\times 23.5 kJ/mol=17,970.6 kJ$

For process 3 :

$Q_3=m\times c_{p,l}\times (T_{final}-T_{initial})$

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:

$Q_3=13000 g\times 2.2J/g^oC\times (0.0-(-33.4))^oC$

$Q_3=955,240 J=955.240 kJ$

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