# Question on heat transferWatch

#1
HI guys,
If you had an enclosed room with insulation and it had lets say objects that had a temperature of around 1800 degrees c would that room reach that temperature? No flames - just heat
Thx
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#2
bump
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4 years ago
#3
Well assuming perfect insulation (not possible)
Still no, because as the objects heated the air in the room they would lose some energy.
With enough objects that have a very high specific heat capacity you could get the room to almost that temp, but overall given enough time everything becomes the same temperature (if there is no interference)
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4 years ago
#4
wait so you have a room and in that room there is an object at 1800c. when you say insulation is it in the room like a vacuum or...

if its in this case the room will be at zero kelins :/
and the objects will remain at 1800
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4 years ago
#5
yh in that case the object tempreture will fall, as it transfers energy to the surroundings as it has the greater energy. so the room will heat up but will not rech that tempreture. the energy the room gains=the energy the objects lose

so by this the room can not equal that tempreture
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#6
(Original post by Thisguy11)
Well assuming perfect insulation (not possible)
Still no, because as the objects heated the air in the room they would lose some energy.
With enough objects that have a very high specific heat capacity you could get the room to almost that temp, but overall given enough time everything becomes the same temperature (if there is no interference)
Thanks for your reply. What if the room was prettu much a metre in length and width and on the walls of the room you had the objecta totalling 1800 degrees,what temperature do you think the room would be at as a minimum? Also may i ask what would be the best insulation to keep temperature high?
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#7
Hi
Thanls for the reply. How long do you think it will take for the temperature to gwt to room temp? In addition what temperature do you think the room will heat to?
Thanks
(Original post by justthatboy)
yh in that case the object tempreture will fall, as it transfers energy to the surroundings as it has the greater energy. so the room will heat up but will not rech that tempreture. the energy the room gains=the energy the objects lose

so by this the room can not equal that tempreture
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4 years ago
#8
(Original post by star80)
Hi
Thanls for the reply. How long do you think it will take for the temperature to gwt to room temp? In addition what temperature do you think the room will heat to?
Thanks
how old are you? it depends on the heat capacity of the air to determine the final tempreture because Q=mc(change in)T. And then make the Final temp the subject by rearranging the formula

how long will it take..mmm never thought about that. to be honest i have no idea!
if i had to say anything it depends on the tempreture gradient, how the colour of the objects( darker objects emit heat better) so i think you could only determine this experimentally
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4 years ago
#9
No, if the whole room (I'm assuming you mean all the matter in the room) were to heat up to 1800, then there had to be some form of energy input. The whole room will eventually reach thermal equilibrium at a temperature lower than 1800 if parts of the room were originally colder than 1800.
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4 years ago
#10
calculate the mass of air mass=density*volume

then plug it into the formula Q=MCT
i then make final tempreture the subject this should give you the tempreture of the air in the room
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#11
(Original post by justthatboy)
how old are you? it depends on the heat capacity of the air to determine the final tempreture because Q=mc(change in)T. And then make the Final temp the subject by rearranging the formula

how long will it take..mmm never thought about that. to be honest i have no idea!
if i had to say anything it depends on the tempreture gradient, how the colour of the objects( darker objects emit heat better) so i think you could only determine this experimentally
(Original post by Protoxylic)
No, if the whole room (I'm assuming you mean all the matter in the room) were to heat up to 1800, then there had to be some form of energy input. The whole room will eventually reach thermal equilibrium at a temperature lower than 1800 if parts of the room were originally colder than 1800.
(Original post by justthatboy)
calculate the mass of air mass=density*volume

then plug it into the formula Q=MCT
i then make final tempreture the subject this should give you the tempreture of the air in the room

Im 17, sorry if my questions seem dumb founded Im not doing A level physics, I were just interested and really curious. So when you talk about thermal equillbruim would you say the temperature could get to like 1000 degrees or? providing that the room was previously at room temperature. The formula seems hard to understand? sorry
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4 years ago
#12
(Original post by star80)
HI guys,
If you had an enclosed room with insulation and it had lets say objects that had a temperature of around 1800 degrees c would that room reach that temperature? No flames - just heat
Thx
So lets use some proper thermodynamic speil.

I'm going to redefine your system to be isolated and kept at constant volume. This means there is no transfer of energy between the system (your room) and the surroundings (everything else outside). So in a sense we have a perfectly insulated room but this also takes into account matter transfer (i.e. what if a window was open and heated air could escape!) and pressure changes (If it's not kept at constant volume, then as the pressure increases due to heating the air in the room then work would be done expanding the system).

So simplifying a bit.... you have only an X amount of heat and it cannot escape the system. It will gradually reach thermal equilibrium within the room. The position of this equilibrium does not depend upon things like the colour of the object (saw someone say that was important before) as all this would change is how quickly equilibrium was reached. The most important factor is the total heat capacity of the system and the heat capacity of the 1800 degree object. We can quite easily make a simple estimation of your problem by multiplying the heat capacity of your object by 1800 to give us the thermal energy in the object. We then need to find the thermal energy of the whole system. So take that and multiply that by the temperature of the system, e.g. room temp perhaps (can you see how many things you've failed to define!)
This gives us the total thermal energy of the system + object. We then divide that through by the total heat capacity (room + object) to give us our equilibrium T.
1
#13
(Original post by JMaydom)
So lets use some proper thermodynamic speil.

I'm going to redefine your system to be isolated and kept at constant volume. This means there is no transfer of energy between the system (your room) and the surroundings (everything else outside). So in a sense we have a perfectly insulated room but this also takes into account matter transfer (i.e. what if a window was open and heated air could escape!) and pressure changes (If it's not kept at constant volume, then as the pressure increases due to heating the air in the room then work would be done expanding the system).

So simplifying a bit.... you have only an X amount of heat and it cannot escape the system. It will gradually reach thermal equilibrium within the room. The position of this equilibrium does not depend upon things like the colour of the object (saw someone say that was important before) as all this would change is how quickly equilibrium was reached. The most important factor is the total heat capacity of the system and the heat capacity of the 1800 degree object. We can quite easily make a simple estimation of your problem by multiplying the heat capacity of your object by 1800 to give us the thermal energy in the object. We then need to find the thermal energy of the whole system. So take that and multiply that by the temperature of the system, e.g. room temp perhaps (can you see how many things you've failed to define!)
This gives us the total thermal energy of the system + object. We then divide that through by the total heat capacity (room + object) to give us our equilibrium T.
Thanks!!!! Just read thru it but atm im dying in pain so will look into tjis! Ty!
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#14
(Original post by JMaydom)
So lets use some proper thermodynamic speil.

I'm going to redefine your system to be isolated and kept at constant volume. This means there is no transfer of energy between the system (your room) and the surroundings (everything else outside). So in a sense we have a perfectly insulated room but this also takes into account matter transfer (i.e. what if a window was open and heated air could escape!) and pressure changes (If it's not kept at constant volume, then as the pressure increases due to heating the air in the room then work would be done expanding the system).

So simplifying a bit.... you have only an X amount of heat and it cannot escape the system. It will gradually reach thermal equilibrium within the room. The position of this equilibrium does not depend upon things like the colour of the object (saw someone say that was important before) as all this would change is how quickly equilibrium was reached. The most important factor is the total heat capacity of the system and the heat capacity of the 1800 degree object. We can quite easily make a simple estimation of your problem by multiplying the heat capacity of your object by 1800 to give us the thermal energy in the object. We then need to find the thermal energy of the whole system. So take that and multiply that by the temperature of the system, e.g. room temp perhaps (can you see how many things you've failed to define!)
This gives us the total thermal energy of the system + object. We then divide that through by the total heat capacity (room + object) to give us our equilibrium T.
Okay, so our object has a heat of 1800 degrees, the heat capacity (not sure) but it's a small confined room circular in shape. So you would multiply them 2 together and then multiply the sum by 23 degrees.
We then divide this by adding the 1800 and the heat capacity of the room and dividing it? This then gives us the max temperature the room will rise to?
How would I work out heat capacity?
Thanks!!!!
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4 years ago
#15
(Original post by star80)
Okay, so our object has a heat of 1800 degrees, the heat capacity (not sure) but it's a small confined room circular in shape. So you would multiply them 2 together and then multiply the sum by 23 degrees.
We then divide this by adding the 1800 and the heat capacity of the room and dividing it? This then gives us the max temperature the room will rise to?
How would I work out heat capacity?
Thanks!!!!

1. Heat capacities are additive not multiplicative.
2. Use kelvin not Celsius.
3. Temperature is not a measure of energy. For that we need the heat capacity and mass of the object too. Go look up the equation.

Not sure what to say after that. Afraid your working makes no sense. I explained the route to the answer above.

We can have a stab at working out the heat capacity of the objects if you define the materials and there sizes. There will be tables of heat capacities for common materials.
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#16
(Original post by JMaydom)

1. Heat capacities are additive not multiplicative.
2. Use kelvin not Celsius.
3. Temperature is not a measure of energy. For that we need the heat capacity and mass of the object too. Go look up the equation.

Not sure what to say after that. Afraid your working makes no sense. I explained the route to the answer above.

We can have a stab at working out the heat capacity of the objects if you define the materials and there sizes. There will be tables of heat capacities for common materials.
Sorry im not doing physics a level thanks for your help anyway
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#17
(Original post by JMaydom)

1. Heat capacities are additive not multiplicative.
2. Use kelvin not Celsius.
3. Temperature is not a measure of energy. For that we need the heat capacity and mass of the object too. Go look up the equation.

Not sure what to say after that. Afraid your working makes no sense. I explained the route to the answer above.

We can have a stab at working out the heat capacity of the objects if you define the materials and there sizes. There will be tables of heat capacities for common materials.
hey! Manged to do it thanks to your help.
used this equation: e = m x c x theta
60 x 2000 x 1000 = 120,000,000 J heat capacity.
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4 years ago
#18
(Original post by star80)
hey! Manged to do it thanks to your help.
used this equation: e = m x c x theta
60 x 2000 x 1000 = 120,000,000 J heat capacity.
Whoah. Heat capacity isn't a measure of energy. Heat capacity measures the amount of energy needed to raise the temperature of the object. So in the equation you wrote down theta is the heat capacity.
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#19
(Original post by JMaydom)
Whoah. Heat capacity isn't a measure of energy. Heat capacity measures the amount of energy needed to raise the temperature of the object. So in the equation you wrote down theta is the heat capacity.
So, sorry im not too sure what you mean. So what does this sum mean?
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4 years ago
#20
(Original post by star80)
So, sorry im not too sure what you mean. So what does this sum mean?
sorry slight mistake by me. Theta is the change in T, c is the heat capacity but my point about heat capacity not being energy stands.
That sum (not sure where you got the data from) tells you the energy change associated with the change in temperature. i.e. how much energy it takes to change the temperature by theta degrees.
If we assume the thermal energy of the material to be zero at 0K (note i need to use kelvin here) then we can find the absolute thermal energy of any object for which we know the mass and the specific heat capacity for. This is the first step we need to complete for all materials in the system for you original problem.
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