How to resolve the heating and cooling of an object?

  • #1
fishspawned
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I am not sure how to proceed. This is just something that i have been thinking about and would love a little help with.

Lets say I have a situation where i am increasing an object's heat energy at some specific rate - for arguments sake, it's 2 degrees every second. If you like we can also rewrite this as an increase of energy of 2 Joules every second. Either will be fine.

let's make it super simple and call it:

dQ/dt [warming] = 2

I am wondering how to incorporate cooling. While this object is being warmed, it is also being cooled.

Newtons 's law of cooling is a function of time with a fixed beginning and ending temperature.

dQ/dt [cooling] = h x A x (T(t) - Tenv)

here h is the transfer coefficient, A is the transfer area, T is the temperature at a certain time, and Tenv is the ambient room temperature.


But what if you are simply starting from room temperature and raising it over time? The rate of cooling is changing as the temperature difference is changing due to the warming.

Is there an appropriate approach to this problem that would determine how much more time it would take for an object to warm to a certain temperature if we also included cooling into the picture? How can i combine the two equations in a sensible way?]

Any help would be greatly appreciated. Thanks
 

Answers and Replies

  • #2
sophiecentaur
Science Advisor
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The sign of the Temperature Difference is what determines the slope of dQ/dt. As with most other formulae, we don't have different equations for heating /cooling, traveling forwards/ backwards, filling / emptying a bucket; the Maths takes care of it all.
 
  • #3
fishspawned
66
15
Perhaps I'm not completely understanding what you are saying, and my apologies for that.
The law of cooling seems to first assume you are at some temperature above Tenv and then will cool down - in the absence of any other heat in or out of the system. (If i am wrong in this, please let me know)

So what I am asking is if there is a different form of the equation if you are starting initially at Tenv but slowly apply heat energy [in which at the same time the object is also experiencing cooling as the temperature is raising.]

Just to be clear - i am suggesting that an outside source is providing additional heat to the system at a constant rate.
 
  • #4
gmax137
Science Advisor
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rate of heat in minus rate of heat out equals rate of change in stored heat. Normally considered as

dQin/dt - dQout/dt = M Cp dT/dt

Where M is the mass of your object, Cp is its specific heat, and T is its temperature (which is a function of time). Keep track of the signs, as @sophiecentaur says, and everything will work out right.
 
  • #5
sophiecentaur
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in which at the same time the object is also experiencing cooling as the temperature is raising
I think I can see your problem. The above statement is implicitly including a source plus a sink for the heat (an extra variable) What your initial equation is telling you is how quickly it will cool down in cold water and it will tell you how quickly it will warm up in hot water (no other heat path involved). That equation of yours is not sufficient for a heater in a room with thin walls (for instance), which is your additional bit.
 

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