Question

Please do part B

TT uriohtent-rid-15131483 1/courses/PHY Long Problem 5, due 2/27/18 aluminum are simultaneously dropped into the water, and the system (insulated are summarized in the table below. A) A massless beaker holds a mass, m.. of water at T-0 C. Masses of iron and comes to equilibrium. The masses, initial temperatures, and specific heats Find the equilibrium temperature, T Substance Water Aluminum Mass mw mw 4 mw Specific heat 0.2 cw 0.1 c Initial T 0 °C 100℃ -40℃ Iron B) Assume m-1 kg and m/2 of ice at-10 °C is added with the iron, Aluminium, and water what is the final temperature? [note: c- cal(g K), Lf 80 cal/g], ce 0.5 cal/(g K). If only part of the ice melts, what fraction of the original ice is left? Assume c for ice is 0.5 cw

Answer 1

A. gi9ven mass leass beaker
mw = mas sof water
Ti = 0 C
specific heat = cw
aluminium mass = mw, T'i = 100 C, specific heat = 0.2cw
iron mas s= 4mw, T"i = -40 C, specific heat = 0.1cw

hecne from conservation of enregy
let final temperature be T
then
mw*cw*(T - Ti) + 4mw*0.1cw(T - T"i) = mw*0.2cw(T'i - T)
(T - Ti) + 4*0.1(T - T"i) = 0.2(T'i - T)
T - Ti + 0.4T - 0.4T"i = 0.2T'i - 0.2T
T = 2.5 C

B. mw = 1 kg
mw/2 ice, Ti = -10 C
cw = 1 cal/(gK)
Lf = 80 cal/g
cice = 0.5 cal/gK = 0.5cw

let mass of ice that melts be m
then

mw*cw(T - 0) + 4mw*0.1cw(T - 0) + mw*0.2cw(T - 0) = 0.5*mw*0.5*cw*(10 - 0) + m*Lf
m = 18.75 grams
fravitno of ice left = (mice - m)/mice = 0.98125