Question

206 40 OA

calculate the weight of the metal B in all the phases for A-60%B at 360 and 100 degrees. If the total weight of the alloy was 183grams.

Please write clearly

Answer 1

Solwy a mass m= 183 grams phase diagram first we draw a B = 60 % given Total A+ 360°c So in the vertical line And also we → So we get → So we drow 1. phases (from point s % (from point 'B) B) % at B=60% = CA draw a horizontal line at T = 360°c point 'A' in (L+ B) phase. point ' ľ and 'B' in pure I and B composition of B in L phase = 55% = 6 Composition of B in B phase. = 92%=45 a fraction of B x = C in Liquid phase (L Phase) CA-Ce - 6o -55 CB-Cc ga-55 the = 0.15625] fraction in ß phase = x 243 - 0.84377 . Now mass of B in given total mass mB= 0.6x 183 gr MB = 109.8 gram 80 At 360°c Mass of B in liquid (L-Phase) = (me) = 26XmB 109.8 * 0.15625 MB) = 17.156 gram

800 a + L Text A L+B B T= 360 °C --x..... a + B 200 410 % B

Mass of B in solid-ß phase a 0.8427x 109.8 are Time)s = 92.644 gram (b) Now At 100°C As in previous case we draw a line at T = 100°C so we get point c'À, B Ae x P ha 4 , C 9 . | Ca - 66 . A ß phase cp = 97% L 97-9 & 80 gration og B in 8-phase - Care - - MB) Hoe = 0.5795 . fraction of B in ß- phose = Gas 97-60 Xp = 0.4205 Mass of B in de phase = xxx mo (mBa = 0.5795 x 109.8 gram 1 (mbla = 46-165 gram mass of B in ß- phase = xBx MB (MB)e = 0.4205.* 109.9 gram Limble = 63.63 grom

890 EMPC a + L Text L + a + B 200 T= 100 °C 1B