Help Meh A 19240 cublc foot room Inltlally has a ndon leval of 8TO plcocuries po
ID: 2890685 • Letter: H
Question
Help Meh
A 19240 cublc foot room Inltlally has a ndon leval of 8TO plcocuries por cublc foot.A vonation system Is Installed that orings in 68 cubic foet ot alr per hour that contains 5 pcoouries per cublc toor, whlle an oqual quantdty of tha well-mixed alr In the room leaves the rcom oach nour. Setup and use a dimerent al equation to cetemnine on long it will take torthe room to reach a sate to breathe level o. 102 pcocuries per cubic toot. (Round your answer to 5 decimal places.) oqual quantity o se a aimaerenica take torExplanation / Answer
let amount of radon in room at time t is r(t)
19240 ft3 of room contains 870 pc/ft3 raden initially
amount of radon initially is r(0)=19240*870
=>r(0)=16738800 picocuries
rate of radon input = 685*5 picocuries per hour
rate of radon output = 685*(r(t))/19240 =685*(r(t))/19240 picocuries per hour
netrate,r'(t)=rate of radon input -rate of radon output
=>r'(t)=685*5 -685*(r(t))/19240
=>r'(t)=685(5 -r(t))/19240
=>r'(t)/(5 -r(t))=685/19240
integrate
=>[r'(t)/(5 -r(t))]dt=(685/19240)dt
=>-ln(5-r(t))=(685/19240)t + c
=>ln(5-r(t))=-(685/19240)t + c
=>(5-r(t))=e-(685/19240)t + c
=>(5-r(t))=Ce-(685/19240)t
=>r(t)=5+Ce-(685/19240)t , c and C are constants
r(0)=16738800
=>5+Ce-(685/19240)*0=16738800
=>5+C=16738800
=>C=16738795
=>r(t)=5+ 16738795e-(685/19240)t
cocentration =(r(t))/19240
cocentration =(5+ 16738795e-(685/19240)t)/19240 picocuries per cubic feet
given safe to breathe level is 102 picocuries per cubic feet
=>(5+ 16738795e-(685/19240)t)/19240 =102
=> e-(685/19240)t=((102*19240)-5)/16738795
=> (685/19240)t=ln(16738795/((102*19240)-5))
=> t=(19240/685)ln(16738795/((102*19240)-5))
=> t=60.20639
it takes 60.20639 hours
please rate if helpful. please comment if you have any doubt
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.