Do hot and cold cancel each other out?

[Fanghur]

There is something I've always been curious about regarding high and low temperatures, namely whether or not hot cancels out cold.

 

Let's say you are inside a freezer that is at a constant temperature of -70 degrees Celsius, also inside this freezer is a burner that is at a constant temperature of +70 degrees Celsius. Now, for this thought experiment we will assume that both the freezer and the burner are kept at their temperatures of +- 70 degrees Celsius. Let's say you were to put your hand on the burner, my question is whether the burner would burn you or if the two temperatures would cancel each other out.

[freezy]

You would most certainly get burnt.

 

Imagine a cube that radiates -70c inside the cube and a sphere inside this cube that radiates out towards the inner surfaces of the cube. Since hot moves to cold, all the heat radiated from the sphere would move towards the cold surfaces. Ideally, these would cancel each other out in the sense that a thermometer would measure 0c in the ambient air. Nonetheless, temperatures would remain hotter closer to the heater and colder near the radiating inner-surfaces of the cube. This is why you would still burn your hand. If you placed your hand halfway between the heat source and the cold source, you should, ideally, feel 0c.

[Reaper]

There is something I've always been curious about regarding high and low temperatures, namely whether or not hot cancels out cold.

 

Let's say you are inside a freezer that is at a constant temperature of -70 degrees Celsius, also inside this freezer is a burner that is at a constant temperature of +70 degrees Celsius. Now, for this thought experiment we will assume that both the freezer and the burner are kept at their temperatures of +- 70 degrees Celsius. Let's say you were to put your hand on the burner, my question is whether the burner would burn you or if the two temperatures would cancel each other out.

 

No! You will not cancel anything out because materials have a specific heat; meaning that it takes energy to heat something up, and a certain amount of energy has to be taken away before it cools back down. Take water for example: it has a specific heat of 4.184 J/(kg*K). This means that you need that much energy to increase it by one degree. And this is only if the water is liquid.

 

Likewise, it takes time before that heat gets radiated away. More specifically, it follows an exponential decay sort of graph on how long it takes for something to cool down to room temperature. And the rate at which this happens also depends on the material; water is very effective in taking away/absorbing heat, which is why it is used as a coolant in many applications. It's effectiveness is also why you never want to be wet when you are lost in a very cold climate.

 

Also, there will never be a situation in which you have one side of a room it is cold and the other it is hot, because heat energy distributes out in all directions. Was it not for ventilation in my room, for example, the room temp would most certainly increase until it is the same temperature as my body.

 

 

If you were to put your hand on a burner, it will get burned. It will not matter if you also put your hand on a tank of liquid nitrogen. And second, you cannot raise the temperature of any part of the body too quickly, otherwise you burst blood vessels.

Edited June 21, 2008 by Reaper

[Mr Skeptic]

There's no such thing as "cold" anyways. If you mix a hot and a less hot thing together, the heat energy of both will be averaged. Since some materials can hold more heat energy per degree of temperature, you can't just average the temperature if they are different materials. Additionally, if a material goes through a phase change (freezing, melting, boiling, condensing, etc) then it will either release or absorb a large amount of heat energy without a change in temperature.

 

However, if you mix the same amount of two identical materials at different temperatures, the resulting temperature would be the average of the two. So mixing one kg of water at 0 degrees, and one kg of water at 100 degrees will result in two kg of water at (very close to) 50 degrees.

[swansont]

Take water for example: it has a specific heat of 4.184 J/(kg*K). This means that you need that much energy to increase it by one degree. And this is only if the water is liquid.

 

J/g*K or kJ/kg*K

[Reaper]

J/g*K or kJ/kg*K

 

Oh, thanks, stupid mistake, got units mixed up .

[Mr Skeptic]

Good thing you're not building a spaceship.

[Reaper]

Good thing you're not building a spaceship.

 

 

Not yet .

[ecoli]

Good thing you're not building a spaceship.

out of water?