Heat vs. Cold Insulation

chienphong

Hi, guys!

(I created the same thread in the Engineering topic nearly a week ago but there has been no response. So I decided to move it here. Thanks.)

A few friends and I just had a discussion regarding heat and cold protection in general. Then came across a question of whether it is more difficult and energy consuming for heat protection or for cold protection, or is it the same. Let's take an example. We have a void cube (with air inside) of a certain peripheral insulation material. Let's say that we somehow adjust the inside temperature of the cube at 10 degrees Celsius and put the cube in a constant air (same type of air as in the cube) environment of 30 degrees Celsius and after 1 hour the temperature inside the cube is 15 degrees Celsius. The question is: if we adjust the inside temperature of the cube at 20 degrees Celsius and put the cube in a constant air (same type of air as in the cube) environment of 0 degree Celsius, what is the temperature inside the cube after 1 hour? Would it be 15 degress Celsius? or higher? or lower?

I would really appreciate the answer and full back up knowledge for that. Thanks a lot!

russ_watters

Welcome to PF.

15C. So the answer to the general question is no, the direction of heat flow doesn't typically matter.

chienphong

Thanks a lot Russ!

However, a little more explanation and few examples would be much appreciated.

I would bring up an example here. The specifications of the air conditioners we are using show that they consume pretty much more power for heating than for cooling. Is it then just a matter of different technologies?

Thanks.

FireStorm000

I don't think that's quite right. never-mind, I used the wrong givens. See newton's law of cooling. You are however correct that the direction doesn't matter.

In short, the rate of transfer of thermal energy should be proportional to the temperature difference between the cube and environment. It is described by a differential equation:
T'(t) = k(T(t)-T_env)
Solving (brush up on your diff eqs here):
T(t)=T_env+[itex]\Delta[/itex]T*e^k*t

To solve the above:
T_env = 30^o; [itex]\Delta[/itex]T = -20^o
T(0)=10^o
T(1hr)=15^o=30^o + (-20^o)e^k(1)
3/4 = e^k
ln .75 = k
k = -.288

And the second part, plugging in k:
T_env = 0^o; [itex]\Delta[/itex]T = 20^o
T(0) = 20^o
T(t) = 0^o + 20^oe^-.288t
T(1) = 20^oe^-.288(1) = 15^o

If you can understand that equation, it should answer all of your questions. In short, direction doesn't matter because there is no privileged frame of reference. Say we were transferring heat between two boxes, one hot, one cold; the rate is not dependent which box we choose as the reference frame (the direction depends on reference frame, but not magnitude)

chienphong

Thanks a lot, FireStorm!

The equation explains it and it is easy to understand. Great help.