Heat Transferred to a Falling Object

[Comeback City]

If I was attempting to calculate the amount of energy transferred as heat to an object free-falling in the atmosphere, is this how I could go about it?...

Work = (Drag Force) x (Displacement through atmosphere) = Energy transferred as Heat

I am attempting to solve a hypothetical question which goes, if a small ice cube is dropped from 30,000 feet in the air, will it melt before it hits the ground? After finding the energy transferred to the ice cube, I would calculate the amount of energy (using specific heat of H₂O) required to both raise the temperature of the ice to 0 degrees and then to change the phase of the ice to liquid water, and compare it to the amount of energy transferred from the fall. Does this process seem to check out?

 

[Charles Link]

No, because the air can heat up rather than the ice cube getting heated.

 

[Comeback City]

No, because the air can heat up rather than the ice cube getting heated.

Thanks for the response!
Is there a way to tell whether or not the air would heat up rather than the ice cube heating up? If not, is there a way to solve this problem?

 

[Charles Link]

There is one additional complicating factor, in that, if there is evaporation that takes place from the liquid converting to gas, the result is 540 calories/gram. In general, the ice cube problem, in my estimation, does not have a simple solution. $\\$ Clearly, without even calculating it, the change in gravitational potential is enough to completely vaporize the ice cube, but whether it results in air getting heated and/or in kinetic energy of the ice cube is another matter. $\\$ Scratch this last statement: Let's compute it: $U=mgh$ , $h=30,000$ ft $\approx 9,000$ meters, $g=9.8$ m/sec^2 , and $m=1$ gram $=.001$ kg . $U=mgh=90$ joules $\approx 20$ calories (1 calorie=4.184 joules). The answer is you can not even melt it with all of the available energy. You need $80$ calories to completely melt it with 100% energy going into heating of the ice cube.

 

[Comeback City]

Thanks for the help!

 

[Chestermiller]

The cube passes through the atmosphere in dropping the 30000 ft, and can exchange heat with the surrounding atmospheric air, since its temperature is going to differ from that of the air. Do you know what the air temperature and density are at 30000 ft, and as a function of altitude? Do you know how to determine the heat transfer coefficient between the air an the object, at the terminal velocity of the object? This is all covered in courses in heat transfer and atmospheric science.

 

[Charles Link]

@Chestermiller Thank you. Yes, that is another part of the analysis that requires careful inspection. It can make a big difference if the atmosphere is at near freezing temperatures, or at temperatures in the $T=20^o$ to $30^o$ C range.

 

[Chestermiller]

@Chestermiller Thank you. Yes, that is another part of the analysis that requires careful inspection. It can make a big difference if the atmosphere is at near freezing temperatures, or at temperatures in the $T=20^o$ to $30^o$ C range.

Actually, at 30000 ft (10 km, base of the stratosphere), the temperatures are on the order of -50 C.