924.8 Frozen Solid (Moscow Phys-Tech)
If the ice does not freeze toofast (which isusually the casewithlakes),
we can assume that thetemperature is distributedlinearly across the ice.
Supposethat the thickness of the ice at a time is Then the heat balance
can be written in the form
where is the meltingtemperature of ice. The left siderepresents the
flow of heat through onesquare meter of ice surface due to the temperature
gradient, and the right side the amount of heat needed tomelt(freeze) an
amount of ice Integrating(S.4.8.1), weobtain
where and are integrationconstants. If weassumethat there is no ice
initially then and we find that the time to freeze
solid is
4.9 Tea in Thermos (Moscow Phys-Tech)
There are two mainsources ofpower dissipation:radiation from thewalls
of the thermos andthermalconductance of the air between the walls. Let
us first estimate theradiativeloss. Thepower radiated from thehotter
inner wall minus the power absorbed from the outer wall is given by (see
Problem4.73)
whereTis the temperature of thetea, is room temperature, and the
Stefan–Boltzmannconstant Initially,
So
The power dissipation due to thethermalconductivity of the air can be
estimatedfrom the factthat, atthatpressure, the meanfree path of the
SOLUTIONS