Question

Calculate the final temperature of a mixture of 350 g of ice, initially at −18.0°C, and 237 g of water, initially at 100.0°C. The answer according to the textbook is 0°C but I can't seem to get to that answer.

Answer 1

So here I again = -8 lost 6 from - 18% to 0°c a tle is melting ф, терът cp = specific heat of ice - 2.108 3/guc mass of ice= 350g Change of temp = ot= (6-(-18) = 18 o Q = mcpot = 350% 2.108 x 7 x 18 % = 1323014 ] © for ice meeting : 2. Q = tf.m pfa po heat of fusion = 334 J/g For ile (required to change from soud to liquid) O = 334*350 P=116,900 ]]

melting of ice from or to final temperature Let's say * .. q- map og (p = 4:184 I/g for water specifir heat m2 350g :q = mlp oto 350 x 4184 x (x-0) = 350X 4.1848 - 1466.64 m from this becil gain by ice cabe 9 gecin = 13280.4 + 116,900 + 1466-43 - from @ ③ e6 cain. - 0 ? gein = 139, 100.4 +1464.26 ice is geining heat hence not water - is losing its heat. q= mcpot m= 237 g cp = 4:134 3796 off too) GT=(x-100) :9 = 239 x 4.186 (0-100) 900 991.60g (7-100) -

Heat geein =-Heat lost. 130,180.47 1464.432 =-991.608 (x - 100) 130,180.4+1464*= -991.608 x + 99160.B पानी - 130180-4 - 99160.8 = -14664991.6082 31019.6 = .-2455.608 x la==126.32 ] 100°C ---19 heat tos heat los Oct 2 (2) 1210 Arm-12°c to -ệcố , 6 @