Question 3. A specific water-based paint (y=62 mN/m, p=1050 kg/m3) has a contact angle of 60 degrees on a certain steel. How large an area can you coat on a horizontal plate with 100 mL? In general, adding surfactant decreases the contact angle. A specific surfactant is supposed to reduce the contact angle to 20 degrees and the surface tension to y=38 mN/m. How much does the coated area increase? For simplicity assume that the rim is an edge rather than being curved.

CAWAQQ The Asker · Chemistry

Transcribed Image Text: 3. A specific water-based paint (y=62 mN/m, p=1050 kg/m3) has a contact angle of 60 degrees on a certain steel. How large an area can you coat on a horizontal plate with 100 mL? In general, adding surfactant decreases the contact angle. A specific surfactant is supposed to reduce the contact angle to 20 degrees and the surface tension to y=38 mN/m. How much does the coated area increase? For simplicity assume that the rim is an edge rather than being curved.
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Transcribed Image Text: 3. A specific water-based paint (y=62 mN/m, p=1050 kg/m3) has a contact angle of 60 degrees on a certain steel. How large an area can you coat on a horizontal plate with 100 mL? In general, adding surfactant decreases the contact angle. A specific surfactant is supposed to reduce the contact angle to 20 degrees and the surface tension to y=38 mN/m. How much does the coated area increase? For simplicity assume that the rim is an edge rather than being curved.
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TL1ZRH

SoL:-We have mar(1-cos theta)=mg=>pi rr^(')(1-omega s theta)=vCgGiven gamma=62mN//mi theta=60^(@) i r= ?{:[v=100mL=100 xx10^(-6)m^(3)],[p=1050kg//m^(3)i*g=9.8m//s^(2)],[r=(10^(-4)xx1050 xx9.8)/(3.14 xx62 xx10^(-3)xx(1-1//2)],[r=10.6m]:}Now Area =pir^(2)=3.14 xx10.6 xx10.6=352.8m^(2)(ii) Given { ... See the full answer