Question

A poorly ventilated commercial kitchen contains 120 kg of air at a temperature of 19°C and and with a relative humidity of 30% at the start of the evening shift.

At the end of the shift the temperature in the kitchen is 28°C and water is condensing on the walls and ceilings.

What is the minimum amount of water vapour that has been added to the air in the kitchen over the course of the shift? (to 2 s.f and in kg)

Answer 1

Solution.

$m_1=120kg;$

$t_1=19^oC;$

$\phi_1=30\%;$

$\rho_{s1}=16.3g/cm^3;$

$t_2=28^oC;$

$\phi_2=100\%;$

$\rho_{s2}=27.2g/cm^3;$

$\Delta m-?;$

$\phi=\dfrac{\rho_a}{\rho_s}\sdot100\%;\implies \rho_a=\dfrac{\phi\rho_s}{100\%};$

$\rho_{a1}=\dfrac{30\%\sdot16.3g/cm^3}{100\%}=4.89g/cm^3=4890kg/m^3;$

$m=\rho V\implies V=\dfrac{m}{\rho}$ ;

$V=\dfrac{120kg}{4890kg/m^3}=0.025m^3;$

$\rho_{a2}=\rho_{s2}=27.2g/cm^3=27200kg/m^3;$

$m_2=\rho_{s2}\sdot V;$

$m_2=27200kg/m^3\sdot0.025m^3=667.5kg;$

$\Delta m=m_2-m_1;$

$\Delta m=667.5kg-120kg=547.5kg;$

Answer: $\Delta m=547.5kg;$