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

Amount of necessary heat transfer, q =
`66.1* 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* 0.200 = 161.6 kg`

Now, from eqn (1):

`q = m(C_(gas)\Delta T + L_(v))`

`q = 161.6(1040(T' - T) + 201* 10^(3))`

`q = 161.6(1040(4 - (- 195.8)) + 201* 10^(3))`

q =
`66.1* 10^(6) J`