Question

Steam at 100° C is added to ice at 0° C. Find the amount of ice melted and the final temperature when the mass of steam is 7.00 g and the mass of ice is 50.0 g.

Answer 1

DO 202 oc Mass of ice - Given: Temperature of steam = 100°c. Temperature of ice Mass of steam = 7.00g 50.0g Solution: We know that Latent heat of fusion of ice = 333 kJ/kg Now, energy required to melt ice is Q = (mice). LA Mice = 50.0g Sox 103 13 Kg Hence, &i= (50x1013) (333) Q, 16.650 KI o Also, the latent heat of vapowisation of water KJ/kg - Lu Hence, the energy given by steam to ice as it condensates, Q (msteam) .Lv Msteam = 7x 103 Kg Q.- (78103912260) = 15.820KI Moreover, the heat given up as steam cores to la malat) = m 7 C(AT) c is the specitic heat = 4186 5 / kg C = 4.186 K5/kic is = 2260 - 7 Steam

Az = [[7.00)x103)/4186)/100-0) 2930.2 J = 2.930 KI Therefore, Total heat pumided by steam to system - Q₂ + Q = (15.820 + 2.930) KJ 18.75 KJ But energy required to melt completence (frem is che & steam To find final Temperature, If Gold = Moe). Le t mcat 16-650° + (50x103) (41186) (TR-0C) | 6: 650 + 0 2pq3 (1) CAT = -15.820+ 0.0293 (TA-100) Hence, all the ice melts Quot = (steam (Lv) + Msteam c

15.820 - 0.0293 Tft 2.93 Gold = - Quot 16:650 +0.20937, +415.820 +0.0293[T. 0:2386 To = 2:1 => Tp = 8.8°C) (final Temperature