Respuesta :
Answer:
Amount of necessary heat transfer, q = [tex]66.1\times 10^{6} J[/tex]
Given:
Volume of liquid Nitrogen, V = 0.200 [tex]m^{3}[/tex]
Density of liquid Nitrogen, [tex]\rho = 808 kg/m^{3}[/tex]
Temperature, [tex]T = - 195.8^{\circ}C[/tex]
Temperature Rise, T' = [tex]4.0^{\circ}C[/tex]
Latent heat of Vaporization, [tex]L_{v} = 201 kJ/kg[/tex]
Specific heat capacity of gas, [tex]C_{gas} = 1040 J/kg^{\circ}C[/tex]
Solution:
Now, calculation of heat transfer required to evaporate liquid nitrogen and raise its temperature is given by:
[tex]q = mC_{gas}\Delta T + mL_{v}[/tex] (1)
Now, mass, m can be given by:
[tex]m = \rho V = 808\times 0.200 = 161.6 kg[/tex]
Now, from eqn (1):
[tex]q = m(C_{gas}\Delta T + L_{v})[/tex]
[tex]q = 161.6(1040(T' - T) + 201\times 10^{3})[/tex]
[tex]q = 161.6(1040(4 - (- 195.8)) + 201\times 10^{3})[/tex]
q = [tex]66.1\times 10^{6} J[/tex]
