UKC

If I have a 3.5kg shoulder of pork wih a specific heat capacity of 2.1kJ/kgK. If it was in a freezer at -15celsius and then left at room temperature at 15degrees, how long should it take till it warms up to room tenperature?

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?

In reply to andy_e:
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).

In reply to andy_e:

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).

In reply to Jon Stewart:

No need to solve differential equations when you have MatLab. I only vaguely recall my degrees maths modules from two years ago!

In reply to Swirly:

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?

In reply to andy_e:

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.

In reply to andy_e:

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.

In reply to andy_e:
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.

In reply to andy_e:
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)

In reply to andy_e:

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
;~))

In reply to andrewmcleod:

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

Simples!

I would be surprised if the temperature gradient across the joint was not important (but could be wrong), and therefore I suspect treating it as having a uniform temperature would be quite inaccurate.

In reply to andrewmcleod:
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)

In reply to James Jackson:
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.

In reply to Jim C:


To explore mathematical curiosities? Nah, it's one of many available joys!

In reply to andy_e: assuming newtons law of cooling T(t)= 15-30exp(-kt) ?

In reply to Jim C: dunk in water..

In reply to James Jackson:

guilty... :P

Defrosting will take of the order of hours, to within an order of magnitude

In reply to James Jackson: logT=-kt+c

In reply to andy_e:

You should defrost in the 'fridge...

In reply to andy_e:
Ask a cook (or Google) rather than a physicist.

In reply to andy_e: AFAIK it depends on the silliness of the belt, the tightness of the trousers and the coefficient of hair floppyness

Andy F

In reply to andy farnell:

Has NOBODY got the answer yet?

In reply to malk:

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.

In reply to James Jackson:

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.

In reply to andy_e:


SPOILERS***
If you want just a simple answer with all the hard work done:
http://webserver.dmt.upm.es/~isidoro/bk3/c11/Heat%20conduction.pdf

Welcome.

In reply to andy_e:
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.

In reply to andy_e:

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!

In reply to Submit to Gravity:


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.

In reply to malk:

Quite. Big bowl, bucket or spare sink, and cover the wrapped joint in water. It'll be ready by tomorrow tea time.

In reply to andrewmcleod:

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.