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almond37 [142]
3 years ago
11

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
Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
6 0

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

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