Question

In some countries liquid nitrogen is used on dairy trucks instead of mechanical refrigerators. A 3 hour delivery trip requires .200 m^3 of liquid nitrogen which has a density of 808 kg/m^3, assuming it is already at its boiling temperature of -195.8C, calculate the heat transfer necessary to evaporate this amount of liquid nitrogen and raise its temperature to 4.0C (Nitrogen: Lv=201 kJ/ kg, cgas = 1040 J/kgC

Answer 1

Answer:

Amount of necessary heat transfer, q = $66.1\times 10^{6} J$  

Given:

Volume of liquid Nitrogen, V = 0.200 $m^{3}$

Density of liquid Nitrogen,  $\rho = 808 kg/m^{3}$

Temperature, $T = - 195.8^{\circ}C$

Temperature Rise, T' =  $4.0^{\circ}C$

Latent heat of Vaporization, $L_{v} = 201 kJ/kg$

Specific heat capacity of gas, $C_{gas} = 1040 J/kg^{\circ}C$

Solution:

Now, calculation of heat transfer required to evaporate liquid nitrogen and raise its temperature is given by:

$q = mC_{gas}\Delta T + mL_{v}$                        (1)

Now, mass, m can be given by:

$m = \rho V = 808\times 0.200 = 161.6 kg$

Now, from eqn (1):

$q = m(C_{gas}\Delta T + L_{v})$  

$q = 161.6(1040(T' - T) + 201\times 10^{3})$  

$q = 161.6(1040(4 - (- 195.8)) + 201\times 10^{3})$  

q = $66.1\times 10^{6} J$