How to resolve the heating and cooling of an object?

In summary, the law of cooling states that the rate of cooling is proportional to the temperature difference and the initial temperature. It also states that the rate of cooling is independent of the heat source and the rate of warming.
  • #1
fishspawned
66
16
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
 
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  • #2
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
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
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
fishspawned said:
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.
 

Related to How to resolve the heating and cooling of an object?

1. How does the temperature of an object affect its heating and cooling?

The temperature of an object directly affects its heating and cooling because the higher the temperature, the more energy it has and the faster it can transfer heat to its surroundings. Similarly, a lower temperature means less energy and a slower rate of heat transfer.

2. What factors influence the rate of heating and cooling of an object?

The rate of heating and cooling of an object is influenced by several factors, including the object's surface area, material composition, and surrounding temperature. Objects with larger surface areas will heat and cool faster due to increased exposure to their surroundings. Different materials also have varying abilities to conduct and retain heat, affecting the rate of heating and cooling. Lastly, the surrounding temperature can either assist or hinder the process, as a higher surrounding temperature can speed up heating while a lower temperature can slow it down.

3. How can I speed up the cooling of an object?

To speed up the cooling of an object, you can increase its surface area by breaking it into smaller pieces or spreading it out. This will increase its exposure to the surrounding environment and allow for faster heat transfer. You can also place the object in a colder environment or use a fan to increase air circulation, which can help dissipate heat more quickly.

4. How can I prevent an object from overheating?

To prevent an object from overheating, you can limit its exposure to heat sources or use insulation to reduce heat transfer. Additionally, you can increase the object's surface area or use materials with lower thermal conductivity to slow down the rate of heating. It is also important to monitor the surrounding temperature and make adjustments as needed to prevent the object from reaching dangerously high temperatures.

5. What is the difference between heating and cooling an object?

Heating and cooling are opposite processes that involve the transfer of heat energy. Heating refers to the process of increasing an object's temperature by transferring heat to it, while cooling involves removing heat from an object to decrease its temperature. Both processes are important for maintaining a stable temperature for various objects, such as food, electronics, and living organisms.

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