Answer:
About 2 units.
Explanation:
We assume that there is no heat dissipating instrument in the room
Total cooling load of the room is defined from the given below equation
[tex]Q_{cooling}[/tex]=[tex]Q_{light}[/tex]+[tex]Q_{people}[/tex]+[tex]Q_{heat gain}[/tex]
where
[tex]Q_{light}[/tex]= 10*100 W =1 KW
[tex]Q_{people}[/tex] = 40*360 KJ/h= 4 KW
[tex]Q_{heat gain}[/tex]=15000KJ/h= 4.17 KW
[tex]Q_{cooling}[/tex]= 1+4+4.17=9.17 KW
the number of air conditioner unit is =[tex]\frac{9.17}{5}[/tex]=1.83
which is approximately 2 units