A perfectly insulating cylindrical glass is partially full of water. At the top
ID: 1526759 • Letter: A
Question
A perfectly insulating cylindrical glass is partially full of water. At the top of the water is a uniform layer of ice. The temperature of the water is 0 degree C, and the temperature of the air above the ice is held fixed at -19.00 degree C. If the initial thickness of the ice layer is 0.0750 m, and the depth of the water below the ice s 0.1050 m, how long, in hours, will it take for the rest of the water to freeze? Assume that there is no transfer of heat through the sides or bottom of the glass, and that the ice layer can slide freely up and down the glass. The latent heat of fusion of water is 3.33 times 10^5 J/kg and the thermal conductivity of ice is 2.180 W/m middot C degree The density of water near freezing is 1000.0 kg/m^3 and the density of ice is 917.00 kg/m^3.Explanation / Answer
growth of ice
amont of heat flowing
Q = K*A*(T2 - T1) *dt/y
K = theraml conducivity of ice
A = area of cross section
T2 = temperature of ice = 0
T1 = temperature of air = -19
dt = time
y = thickness of ice
for ice
heat lost Q = m*L
m = mass = D*A*dy
D = density
A= area of cross section
dy = thickness of ice formed
K*A*(T2-T1)*dt /y = D*A*dy*L
dy = K*(T2-T1)*dt/(D*L*y)
y*dy = k*(T2-T1)*dt/(D*L)
y^2/2 = k*(T2-T1)*t/(D*L) from y = y1 to y2
t = (y2^2-y1^2)*D*L/(2*k*(T2-T1)
y1 = 0.075 m
y2 = 0.075+0.105 = 0.18
(T2-T1) = 19
dt = time = ?
D = 1000 kg/m^3
L = 3.33*10^5
K = 2.18 W/mc
t = (y2^2-y1^2)*D*L/(2*k*(T2-T1)
t = (0.18^2-0.075^2)*1000*3.33*10^5/(2*2.18*19)
t = 107630 s
1h = 3600s
t = 107630/(3600)
t = 29.89 hours <<<-----answer
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