How Does Ice Form in a Tank of Water Left Outdoors in Cold Weather?

[zferic28]

1) If a tank of water is left outdoors in very cold weather, a sheet of ice will form on the water suface. Assume that, sa the water freezes on the underside of the ice sheet, the heat of fusion is conducted through the ice. For parts (c) and (d) assume that the upper surface of the ice sheet is at -10C and that the water below the ice is at 0C. Assume that there is no heat transfer through the walls or the bottom of the tank.

a) When the air temperature is below 0C, the water at the surface freezes to form an ice sheet. Why doesn't the freezing occur through the entire volume of the tank?

b) Show that the thichness of the ice sheet formed on the water surface is proportional to the square root of the time elapsed since the freezing started.

c) Calculate the thickness of the sheet that will form in one day.

d) If the tank is 50cm drrp, how many days will it take to freeze all the water in the tank?

e) If this were a lake, unifrmly 10 m deep, how many days would it take to freeze all the water in the lake? Is this likely to occur?

3) Consider a poor lost soul walking at 5.0 km/h on a hot day in the desert, wearing only a bathing suit. this person's skin temperature tends to rise due to four mechanisms:
(i) Energy is generated by the metabolic reactions in the body at the rate of 280W, and almost all of this energy is converted to heat flows to the skin;
(ii) Heat is delivered to the skin by convection from the outside air at a rate equal k'A_skin_(T_air_-T_skin_), where k' is 54 J/h-C-m^2, the exposed skin area is A_skin_ is 1.5 m^2, the air temperature T_air_ is 47C, and the skin temperature T_skin_ is 36C
(iii) The skin absorbs radiant energy from the sun at the rate of 1400 W/m^2.
(IV) The skin absorbs radiant energy from the environment, which has the temperature 47C.

a) Assume the emissivity of the skin is e = 1 and the skin temperature is initially 36C Show that the rate in watts at which the person's skin is heated by each of the last three mechanisms is:
(ii) 0.248W
(iii) 2.10 x 10^3 W
(iV) 893w

B) At what rate (in L/h) must perspiration evaporate from this person's skin to maintain a constant skin T.
c) Supposed instead that the person is protected by light colored clothing (e=0), so that the expoed skin area is only 0.45 m^2 What rate of perspiration is required now.

If somebody could tell me which formula to use for each step and set up the equation it would be much appreciated. I added this course about a week late and this HW is due today and I havn't even been to any classes. Absolutely any help or guidances is highly appreciated. I am starting the chapter right after I hit the submit button and should have some sense in a couple of hours. But please, post any suggestions.

Thanks, Z

 

[Jhero]

a) The freezing only occurs at the surface of the water because the air temperature is below 0C and the water at the bottom of the tank is still at 0C. This creates a temperature gradient where the surface of the water is colder and thus more likely to freeze first.

b) To show that the thickness of the ice sheet is proportional to the square root of time, we can use the equation for heat conduction through a material:

Q/t = kA(T1-T2)/d

Where Q is the heat transferred, t is time, k is the thermal conductivity of the material, A is the surface area, T1 and T2 are the temperatures on either side of the material, and d is the thickness of the material.

Since we are assuming no heat transfer through the walls or bottom of the tank, we can rearrange the equation to solve for d:

d = kA(T1-T2)(t/Q)

Since the heat of fusion is conducted through the ice sheet, we can replace Q with the heat of fusion (Q = mL) where m is the mass of the ice sheet and L is the latent heat of fusion for water. We can also assume that A remains constant and that T1 is at -10C and T2 is at 0C.

d = kA(-10C-0C)(t/mL)

Since we are looking for the thickness of the ice sheet, we can replace m with the density of ice (m = pV) where p is the density of ice and V is the volume of the ice sheet. We can also assume that the density of ice is constant.

d = kpA(-10C-0C)(t/pV)

Now we can solve for the thickness (d) at a specific time (t):

d = sqrt(kpA(-10C-0C)t/pV)

Since we are only interested in the relationship between thickness and time, we can ignore the constants (k, p, A, -10C, 0C) and simplify the equation to:

d = sqrt(t/V)

c) To calculate the thickness of the ice sheet that will form in one day, we can plug in 1 day (24 hours) for t and the volume of the tank for V. Let's assume the tank is a perfect cube with dimensions of 50cm x 50cm x 50cm.

d = sqrt(24 hours/(50cm