/ Physics question about defrosting meat
So far I have figured it needs 220kJ of energy input to warm up, but I'm stuck at a heating rate. Presumably this is related to the temperature difference of the meat and the room, and the rate decreases in some linear (or at least predictable) fashion as the two temperatures converge.
Any one with a better memory of physics able to help?
There is a thing called the heat equation - it's a differential equation (which is what you'd expect as the rate of change of temperature at any one point in the joint is going to depend on the difference in temperature at the adjacent points).
You're mostly concerned about the size of the thing, the crucial question being when is the middle at the same temperature as everything else.
So, it's reasonably tricky maths (well beyond A-level).
This is one of those things when you're better off asking your mum than trying to consider the physics!
For a start the shc will change when the water melts (along with the latent heat of melting) and the room isn't a closed system (unless you have some crazy insulation).
No need to solve differential equations when you have MatLab. I only vaguely recall my degrees maths modules from two years ago!
The room can be considered large enough in relation to the meat to be closed? And the shc is 2.6 above freezing and 1.4 kJ/kgK. Yeah?
Yeah I would approximate your joint to a sphere and just consider T(x,t), seeing when T gets close enough to 15C at the origin. I think that the latent heat and change of shc might throw a bit of a spanner in though. I really wouldn't spend much more time thinking about this - if you didn't know where to start, you're definitely not going to finish.
I don't know the exact figures for pork but I'd approximate to water which is 4.2 as liquid and 2.1 as a solid.
The surface area is going to have a decent impact on heating rate too.
As a physicist, this is one of those complicated things best worked out either empirically (by testing) or numerically (which would itself need to be informed by testing).
The problem is that it depends on things like the thermal conductivity of the meat at different temperatures; everything will be very complicated and even if you model the joint as a sphere (to make things a bit simpler), and assume a uniform thermal conductivity, you would still want to work it out numerically.
My knowledge is Just to Higher Physics
But if you put it in a microwave at 'defrost' ( that is a technical term) , it will be done quicker than without putting it in a microwave. ( and you also need to turn it on)
My youngest daughter( who has a degree in Maths and Physics) agrees.
(Like father like daughter)
You also need to factor in the cooling effect of evaporation.
The humidity of the room will also affect the rate of evaporation.
It is also easier to calculate if the meat is on a conveyor belt rather than a turntable
Or by making further approximations. Assume that the object can be modelled as uniform, with a uniform temperature at all time (valid for a suitably slow process), and that the process is time-reversible (it is), and then apply Newton's law of cooling:
dT / dt = kT
where T is temperature, and k can be calculated from a few assumptions from the object (relative SHCs of the meat, the air, etc).
This equation is trivial to solve, by the way. T here is really (T(object) - T(environment)), so you just want to work out k (dimensional analysis will help you here to work out the ingredients (heh) required), integrate in t with the initial condition (t = 0, T = t(object)), then set T = 0 to give you t when the objects are the same temperature.
Works out (up to a minus sign) as:
t = kT
where T is defined as above (temperature difference). The trick is in calculating k. That is left as an exercise for the reader ;-)
If you think that's a stab in the dark, you should see what astrophysicists get up to.
(P.S. I agree. Would be interesting to see the full numerical solutions vs that probably crap approximation)
Ok my daughter and I have reviewed the problem, and after reading ewar's conveyer comments( and ignoring them) we have concluded that all these fancy calculations are already on a chip IN our microwave, so :-
Step 1 - select Defrost
Step 2 - select 'meat joint'
step 3- enter joint - Weight
step 3- remove any wrapping
Step -5 - switch on microwave.
Step -6 listen for a 'ting' sound.
Life is too short.
To explore mathematical curiosities? Nah, it's one of many available joys!
Defrosting will take of the order of hours, to within an order of magnitude :)
You should defrost in the 'fridge...
Ask a cook (or Google) rather than a physicist.
Has NOBODY got the answer yet?
Gah midnight typing. Indeedy.
The model clearly breaks down as setting T = 0 doesn't work. Need to take the limit as T->0 I suppose.
All temperatures in any equation where you are using absolute, not relative temperatures, will need to use an absolute temperature scale (e.g. Kelvin). In such a scale, T=0 is a silly place anyway.
If you want just a simple answer with all the hard work done:
From someone who's had the worst week of their life and is in for s pretty f***end up Xmas, thank you so much for this thread which Has had me in stitches. :)
It's a bit more complicated - basically a three stage sequential process. First you have to heat the pork up to 0C (mass x 15 x specific heat capacity), then you have to melt all the water in it (mass of water x specific latent heat of fusion of water), then you have to heat it all to 15C (mass x 15 x specific heat capacity). I would make the simplifying assumption that the room is very large and the air is circulating so you can always have air at 15C in contact with the pork ... then set up the equations for each stage!
I recall being in the chip shop in Spean Bridge in winter, a few years back, we had a coffee to warm us up whilst we waited for the fish to cook, the woman serving had forgotten to put the milk jug in the fridge, and left it on the counter, the milk was frozen solid, she put it in the fridge again , presumably to defrost.
Quite. Big bowl, bucket or spare sink, and cover the wrapped joint in water. It'll be ready by tomorrow tea time.
Nah, it still breaks down. Now it's not 1am, the complete equation is:
dT(t) / dt = -k(T(t) - Ta)
where Ta is the atmospheric temperature. Solving this gives:
T(t) = Ta + (T0 - Ta) * exp(-kt)
where T0 is the initial temperature of the object. Solving for fully defrosted implies T(t) = Ta, which still results in a log(0). Makes sense as it's an asymptotic process.
Elsewhere on the site
Make the most of this months HALF PRICE OFFER on the Five Ten Guide Tennie Mid!! Designed as a hybrid approach and... Read more
The British climbing scene is very exciting at the moment. It is quite clear that as a sport it is developing at a rapid rate and... Read more
Hot Aches Productions premiered their latest film Redemption: The James Pearson Story at Kendal Mountain Festival on... Read more
The Christmas Gift Guide at Outside.co.uk Check out our top selection of Christmas Gift Ideas for climbers,... Read more
2012 saw the release of the beautiful first volume of definitive Yorkshire Gritstone climbing, produced by the YMC with Robin... Read more