Sol. The thermos which loses less heat in a given time will be the better one. Exchange of heat takes place between the thermos and the surrounding atmosphere through the sidewalls and the ends of the thermos. If the capacity and the height are the same, the sectional area, and so the end area, is also the same. But the surface area of the side walls is not the same; the surface area of the sides of the cylindrical one is less than that of the square thermos. Therefore the thermos with the circular cross-section loses less heat than that with the square cross-section, and it is accordingly better.

For a strict proof of the fact that, given equal sectional areas, the surface area of the side walls of the circular thermos is less than that of the square section, the following calculation may be carried out.

The equality of the end areas (Fig.) gives us that

= ,

Hence

d 2 = d 1 ,

The surface area of the side walls of the cylindrical thermos

S 1 = π d 1 h.

The surface area of the side walls of the square thermos

S 2 = 4d 2 h = 4hd 1 = d 1 h .

Comparing these results we find that

S 1 < S 2 .