MadSci Network: Physics

Re: How fast does ice melt in air?

Date: Fri Oct 5 14:41:03 2001
Posted By: Joseph Weeks, President
Area of science: Physics
ID: 1000192872.Ph

What a great question.  The rate that the block of ice will melt will 
depend upon two forms of heat transfer, convection and conduction.  Air 
surrounding the block of ice will transfer heat to the ice through the 
process of convection.  If the block is sitting on a table or other solid 
surface, heat will be transferred to the block of ice through conduction.

Since the size of the block is specified in pounds, we need to be able to 
convert between english and SI units, even though most are taught SI units 

The first question is "how big is a 300 pound block of ice?"  We need to 
know this because it will help us to estimate the surface area of the 
block.  The density of ice is 0.9160 grams/cubic centimeter (see for a brief discussion of ice 
density and a phase diagram).  So a 300 pound block of ice weighs 135,900 
grams.  If we divide grams by density, we learn that the block occupies 
about 148362 cubic centimeters.  If we assume that the block of ice is a 
cube, then taking the cube root of 148362 tell us that the block is about 
53 centimeters on a side.  Converting to english units, the block is about 
20.8 inches on a side.

The amount of heat that is transferred to the surfaces of the block of ice 
exposed to air is controlled by the equation: q=h*A*(T2-T1), where q is the 
amount of heat transferred, h is the heat transfer coefficient, A is the 
surface area of the block, and T2-T1 is the difference between the average 
air temperature and the temperature of the block of ice.  There are more 
precise calculus equations, but the equations presented here are probably 
good enough (plus it is hard writing all those funny symbols in a text 

If the bottom of the block of ice is resting on a solid surface, the rate 
of heat conduction to the bottom block is controlled by the equation 
q=k*A*(T2-T1/x2-x1) where k is the thermal conductivity, A is the surface 
area, T2-T1 is the change in temperature measured across a distance x2-x1. 
 Since we have absolutely no information about whether the block is on a 
metal or wood table (which have big differences in conductivity), let's 
just simplify the problem and assume that the rate of conduction through 
the bottom of the block is the same as the rate of convection over the 
other 5 surfaces of the block.

The questions now are what is the average air temperature around the block 
and what is the average heat transfer coefficient.  Let us assume a room 
temperature of 22C.  For an object sitting in air without a fan or other 
forced convection, the average heat transfer coefficient varies from 1 to 5 
Btu/hr square foot F or 6 to 30 W/m^2K.  A good average value is probably 2 
Btu/hr or about 12 W/m^2K.  

Let's consider what happens in the first hour that the block begins to 
melt.  The block is 53 centimeters on each side or has a total area of 
53*53*6=16854cm^2 or 1.68m^2.  Lets plug all of this information into our 
heat transfer equation. q=12 w/m^2 K * 1.68m^2 * (22-0)K or q=444 watts.  
As the ice melts, the water drips away so we will assume that all of the 
heat goes to melt the ice.

The heat of fusion (the amount of heat necessary to transform ice to water) 
is 80 cal/gram or 333 joules/gram.  Now if you remember that a joule is a 
watt per second. 1 watt=1 joule/second so the total amount of heat 
transferred is equal to 444 watts * 3600 seconds/hour or a total of 1.598 
million joules.  Divide by 333 joules/gram of water, we find that in the 
first hour 4,800 grams of water have melted.  Now the block weighs only 
135900-4800=131,100 grams.

Now all you have to do is to figure out how big the block is now, and 
repeat the calculation, using a slightly smaller block to calculate surface 
area.  Just to get you started, the block is now about 52.3 centimeters on 
each side.  Just keep going and you will eventually get to an answer that 
should be a lot better than a guess.

Of course there are a lot of assumptions that have gone into this 
calculation.  It also should help you to understand how a small error in 
something like the heat transfer coefficient or room temperature 
measurement can make a big difference over a long time.  Good luck.

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