Question

Canvas XCO Composition as 0 100 20 60 80 40 1 327°C 600 300 Liquid 500 232C a + L 200 400 183°C 8 Temperature (°C) Temperature (7) 18.3 61.9 97.8 300 100 4+ 200 100 0 O 20 40 60 80 100 (Pb) Composition (wt% Sn) (Sn) 3. A Pb-Sn phase diagram is given Answer the following questions. There is no need to re-draw the diagram in your solution sheet. (a) What is maximum solubility of Sn in Pb (1 point) (b) A Pb-Sn alloy with 30 wt% Sn is slowly cooled from a completely molten state. At approximately what temperature would the first solid nuclei be observed? Describe in one or two sentences how you arrived at this value. (3 points) (c) How much Sn in wt% is present in the eutectic Pb-Sn alloy? (1 point) (d) 100 g of Pb-Sn alloy with 30 wt% Sn is slowly cooled to 100 degrees Celsius. What are the phases present in this material? Approximately how much (in grams) of each phase is present in this alloy? (1 + 4 points) 12pt Paragraph B I U AV *** hp

Answer 1

Solution 1324c 600 300 Liquid 250 500 Temperature (°C) Oth Eutectic point 2007 B+L Temperature (°F) 183 400 61.9 978 18-3 300 190 atß 200 1 100 60 40 30 80 20 100 (sn) comparition (wt% sn) (@ 18-3 wt% of sn. → masumum salubelety of sn mpb is (b) → → approsumate temperature is 250C when we coal a pb-sn alloy with 30 wilsn from a completely malton state we get the phase (Q+L) where we observe fret cald nuclei. -> And the approxumate temperature when we get (+4) phase during coaling at 30 Wtt. Sn is 250°C 250c L ath 30 wts Sn

at Eutectic Pb-Sn alloy 61.9 wit y of Sn is present (d) when Pb-sn alloy with 30wt% Sn is slowly coaled to 100°c then there is only two phases are present a and ß. Amount of alloy ( Pb-sn) = 100 g and of sn in 30 wty sn means there is 30 gram loo grom of Pb-Sm alloy. And because here a and ß only present then there to g of pb in 100 gram of Pb-sn alloy So here = 30 gram amount of sn in 100g Pb-sn alloy amount of Pb in loog Pb-sn alloy =gram Answer