A practical question

I only have numbers for the domestic units handy, which is a nuisance.

R values for styrofoam run R = 4 to 7 usually. Use R = 4

1/R = k which in the domestic system, means 1/4 BTU per hour per sq ft exposed at 1 inch thick for 1 deg F difference

Let's start with some conversions....

1 BTU = 1055 joules 1055 joules per hour = 1055 / 60X60 = 0.293 watts

1 square foot area = 12 X 12 / 39.37 X 39.37 = 0.0929 sq meters

1 degF = 5/9 deg C = 0.555 degC

So the thermal resistance in question is

0.25 X 0.293W per 0.0929 sq meters per 2.5 cm thickness at 0.555 degC delta (Phew!)

We have 0.77 m^2 area , 5 cm = 2 inch thickness and 28 degF = 15.6 degC delta and time = 24 hours = 86400 seconds. The heat flow in terms of power transfer - watts is

0.25 X ( 0.77 / 0.0929 ) X 0.293 W X 15.6 degC / (0.555 degC/degF X 2 in thickness ) = 8.53 watts during 24 hours.

How to provide this heat flow rate? Thinking about ice to ice water, that's 80 cal/gm or 80 X 4.2 joules / gm of melt = 336 joules /gm

8.53 watts for 24 hours = 8.53 X 60 X 60 X 24 = 737000 joules

Thats 737000 / 336 gms of melting ice = 2.2 kg of ice.

[If this were my chore, I would double check these figures!!]

You're welcome

Brian W

This doesn't count as a recheck even, but I notice I used a delta of

28 degF WRONG! s/b 28degC = 28 X 9/5 degF = 50degF That's much more ice - unless there are other numeric zits in there..... [that's what hapeens when you mess round converting units.]

Brian W