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Question 8 A tank of water has been outdoors in cold weather, and a slab of ice 3.9 cm thick has formed on its surface (see the figure). The air above the ice is at 13°C. Calculate the rate of formation of ice (in centimeters per hour) on the ice slab. Take the thermal conductivity of ice to be 0.0040 cal/s cm-C°, its density to be 0.92 g/cm3, and its latent heat of fusion to be 333 kJ/kg. Assume no energy transfer through the tank walls or bottom. Air Ice Water Number Units Question 10 The temperature of a 0.712 kg cube of ice is decreased to -153°C. Then energy is gradually transferred to the cube as heat while it is otherwise thermally isolated from its environment. The total transfer is 0.700 MJ. Assume the value of cice= 2220 J/kg.K is valid for temperatures from -153°C to 0°C. What is the final temperature of the water? Number Units the tolerance is +/-2 %

Answer 1

Q8.

Rate of Heat transfer through the ice slab is given by:

dQ/dt = -k*A*dT/dh

Given that k = thermal conductivity of ice = 0.0040 cal/sec-cm-C

dT = Change in temperature = -13 - 0 = -13 C

dh = thickness of ice slab = 3.9 cm

A = Area of Slab

dh/dt = rate of formation of ice = ?

Using above values:

dQ/dt = -k*A*dT/dh = -0.0040*A*(-13)/3.9

dQ/dt = 0.01333*A cal/sec

Now rate of ice formation will be:

dQ/dt = m*Lf/dt

Lf = Latent heat of fusion = 333 kJ/kg = 80 cal/gm (Since 1 cal = 4.184 J)

m = mass of ice = density*Volume = p*V = p*A*dh

dh/dt = (dQ/dt)/(p*A*Lf)

p = density of water = 0.90 gm/cm^3

Using known values:

dh/dt = (0.01333*A cal/sec)/((80 cal/gm)*(0.90 gm/cm^3)*A)

dh/dt = 0.01333/(80*0.90) = 0.000185 cm/sec

1 hr = 3600 sec, So

dh/dt = (0.000185 cm/sec)*(3600 sec/1 hr) =

dh/dt = 0.666 cm/hr

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