A hollow copper ball and a solid copper ball of the same size are heated to 100 degree Celsius. They are then placed on a sheet of ice.After 10 minutes, one of tha balls passes through the sheet of ice. Which ball is most likely to be the one that can remain on the sheet of ice? Briefy explain.THX!
In order to go through the ice sheet the two spheres of equal radius should melt the same quantity of ice giving up their internal energy which is propostional with mass. The conclusion is that the hollow copper ball which has less internal energy will remain on the sheet of ice.
The solid one will melt through faster, as all respondents except one have correctly noted. Be careful of the explanation offered by Johnsson W -- it contains a huge oversight and is an example of a little knowledge being dangerous. He states that the LHC of air is greater than copper -- probably true, but I’m not going to bother looking it up because it doesn’t matter. The problem stated the two balls were the SAME SIZE. While it does NOT state that he hollow ball contains air (why not argon or fluorine or a vacuum?), the hollow ball will be much lighter. The difference in a 10-inch ball, say, would be KILOS of copper versus grams of gas. Thus, the solid ball at 100C is going to hold a lot more hear. Johnsson W would be correct only if the two balls weighed the same.
The hollow one. This question relates to the latent heat capacity of materials. It is known that the LHC of copper is much less than that of air. The hollow ball contains mostly air while the solid one contains mostly copper, of course. Now for both balls to reach 100C, the hollow ball needs to absorb a lot more heat than the solid one, because the energy required to change the temperature of air by one unit is much more than copper. So when the two balls are placed on thin ice, the hollow ball releases more heat to melt the ice, while the solid one does not release as much. Therefore, the hollow one is the first to pass through the ice sheet.
