How long does it take for a block of ice that is 3'0" on each side to melt?

What is the room temperature?

Is the block on a surface or suspended?

Will the block be sitting in the runoff?

What is the heat-source?

There are a variety of variables which one needs to consider before answering.

quote:

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Originally posted by MrSensitive:
What is the room temperature?

Is the block on a surface or suspended?

Will the block be sitting in the runoff?

What is the heat-source?

There are a variety of variables which one needs to consider before answering.

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Thanks for telling me! I hope this is enough information because it's all I have to go on right now.

The block of ice is outdoors, but in the shade (no direct sunlight), sitting on a small wooden table 3' x 3' x 1" thick. The ambient air condition is 95 degrees F, with a relative humidity of 50%. There is a 5 mph wind. These conditions will be constant until the block of ice has melted.

Do you need anything else?

Glad to see you are finally back here where you belong. This place is looking better every day. Welcome. Now I'll never have to embarrass myself by trying to answer chemistty questions.

Hello Frankvan,
Glad to see you're still spending your free time answering questions... The fact that you're here means I don't have to embarrass myself by answering the electrical questions. I guess it would be nice if we'd actually answer this question (sorry Mr. Sensitive):


q = (m)*(Hf)


m the mass is equal to (3*3*1/12)(57.7 lbs/ft3) = 43.3 lbs and Hf, the heat of fusion, is equal to 143.3 BTU/lb. therefore q = (43.3)*(143.3) = 6201 BTU's


and q/hr = U*(A)*(deltaT)


delta T is the easiest one so let's get that one out of the way first delta T = (95 -32) = 63


A the exposed surface area is the next easiest to determine/estimate, based on the dimensions you've given you can assume its a wall, or rather two walls one made of ice and the other of wood, with an exposed surface area of roughly 9 ft2.


OK now the hard part determining U (the overall heat transfer coefficient).
U = 1/[1/hi + xa/ka + xb/kb + 1/ho]


where hi and ho are the inside and outside "resistance to heat transfer" terms. For the conditions you've given (a steady 5 mph wind) they should be somewhere between 10 and 20 so let's assume they're both equal to 15.
Xa and xb are the thicknesses (in feet) of the ice and the wood respectively. You didn't give the thickness of the wood so I'm going to assume its 1/6 of a foot (2 inches). ka and kb are the coefficients of heat transfer for the ice and the wood. ka = 1.3 BTU ft/(hr ft2 F) and kb = 1.5 BTU ft/(hr ft2 F). So solving for U:


U = 1/[1/15 + (1/12)/(1.3) + (1/6)/(1.5) + 1/15] = 1/[.067 + .064 + .111 + .067] = 1/.309 = 3.24 BTU/hr ft2 F


q/hr = (3.24 BTU's/hr ft2 F)*(9 ft2)(63 F) = 1837 BTU's/hr and finally:


hours = (6201 BTU's)/ (1837 BTU's/hr) = 3.4 hours (please note this is a very rough estimate... its accuracy is dependent on the assumptions made about hi and ho). If we had assumed hi and ho were 10 not 15 we'll get an answer of 4 hours and if had assumed they were equal to 20 we'd get 3 hours.

[This message was edited by cibby on 06-13-02 at 04:51 PM.]

In my answer to this question I wrote: "where hi and ho are the inside and outside "resistance to heat transfer" terms.


I should have written that 1/hi and 1/ho are the inside and outside resistance to heat transfer terms.


In fact hi and ho are individual heat transfer coefficients. The higher each one of these is the greater the value of U (the overall heat transfer coefficient), and the faster the heat is transferred. Typically for natural convection h will vary from 1 to 5 and for flowing gases it will vary from 2 to 50.

[This message was edited by cibby on 06-14-02 at 11:49 AM.]

Cibby I wish you lived in Texas so I could shake your hand personally. You put a lot of work into this.

Does the relative humidity make any difference, or is this made up for by your U factor? It seems like higher humidity air would make the ice melt faster.

Y'all are funny with the pictures and stuff.

hi and ho are basically engineering "fudge factors" so they're meant to take into account the fact that the air moving past the block of ice contains a normal amount of water vapor. However what I would say is:

1. With no wind speed the humidity would have some effect on the melt time.

2. With high wind speeds humidity would make virtually no difference, and the rate of heat transfer would be controlled by the thermal conductivity of the ice, its shape, and its thickness.