A rectangular block of solid carbon floats at the interace of two immisicible li
ID: 734027 • Letter: A
Question
A rectangular block of solid carbon floats at the interace of two immisicible liquids. The bottom liquid is a relatively heavy lubricating oil, and the tip liquid is water. Of the total block, 54.2% is immersed in the oil and the balance is in the water. In a separate experiment, an empty flask is weighed, 35.3 cm3 of the lubricating oil is poured into the flask and the flask is reweighed. If the scale reading was 124.8 g in the first weighing, what would it be in the second weighing? (recall archimedes principle)Explanation / Answer
volume of oil=0.542volume of block ?(block) x V(block)=?(oil) x (0.542Vb) ?(oil)=?(b) x V(b)/0.542Vb=2.26/0.542= 4.17 g/cm^3 m(oil)=? x V(poured) = 4.17 x 35.3 = 147.2g m(total)=m(o) + m(f) = 147.2g +124.8g = 272g