thermal equillibrium Watch

MSB47
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can someone help me on how to tackle this question im not sure about thermal equillibrium idea behind it. The answer is 82 by the way... Thanks
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Actaeon
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Two bodies are in thermal equilibrium when they are at the same temperature.
In cases like this, we assume all the energy remains inside the system, so a certain quantity of heat energy Q will flow from the water to the cup until both are at the same final temperature (which is what you're trying to find).

So to solve it, we need to form two heat capacity equations; one for the water and one for the cup. As Q is the same for both equations, we can equate them, and solve for our unknown final temperature. Hope this helps a bit
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MSB47
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(Original post by Actaeon)
Two bodies are in thermal equilibrium when they are at the same temperature.
In cases like this, we assume all the energy remains inside the system, so a certain quantity of heat energy Q will flow from the water to the cup until both are at the same final temperature (which is what you're trying to find).

So to solve it, we need to form two heat capacity equations; one for the water and one for the cup. As Q is the same for both equations, we can equate them, and solve for our unknown final temperature. Hope this helps a bit
So the water will lose energy while the cup gains energy until both have the same temperature?

So would you use Q=mc(delta)T? I'm still slightly confused because the mass of the cup is not known
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Actaeon
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(Original post by MSB47)
So the water will lose energy while the cup gains energy until both have the same temperature?

So would you use Q=mc(delta)T? I'm still slightly confused because the mass of the cup is not known
That's right

The heat capacity of the cup is the quantity of thermal energy needed to raise the temperature of the entire cup by one kelvin.

The specific heat capacity is the heat capacity per unit mass.

So for the water, you do use \Delta Q = mc \Delta \theta .
But for the cup, you're not told the specific heat capacity, you're told the heat capacity. And so you need the equation for heat capacity, not for specific heat capacity - it's very simple, it's just  \Delta Q = C\Delta\theta where C is the heat capacity of the cup.
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MSB47
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(Original post by Actaeon)
That's right

The heat capacity of the cup is the quantity of thermal energy needed to raise the temperature of the entire cup by one kelvin.

The specific heat capacity is the heat capacity per unit mass.

So for the water, you do use \Delta Q = mc \Delta \theta .
But for the cup, you're not told the specific heat capacity, you're told the heat capacity. And so you need the equation for heat capacity, not for specific heat capacity - it's very simple, it's just  \Delta Q = C\Delta\theta where C is the heat capacity of the cup.
ive equated both equations however, delta T will just cancel as they would be the same values too and im not sure Q would be the same because its only the internal energies that are equal unless the thermal energy of the water and cup is the internal energy??

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Joinedup
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(Original post by MSB47)
ive equated both equations however, delta T will just cancel as they would be the same values too and im not sure Q would be the same because its only the internal energies that are equal unless the thermal energy of the water and cup is the internal energy??

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Delta T is a difference in temperature not the equilibrium temperature.

What's going to happen here is that the cup is going to increase in temperature and the water is going to decrease in temperature until cup and water are at the same temperature... Teqm

...and we're going to obey conservation of energy.

ΔTcup=Teqm-20 (we can see that it's going to be positive if Teqm is greater than 20)

ΔTwater=Teqm-90 (which is going to be negative if Teqm is less than 90)

so we can write

ΔQcup = C ΔTcup
ΔQwater=m c ΔTwater

ΔQwater=-ΔQcup (the reduction of the energy in the water equals the increase in energy of the cup)

and solve this


-----
alternative method (my 'absolute' favourite)

convert temperatures to Kelvin (add 273.15)
multiply the starting temp of the cup and the water by the appropriate heat capacity to get the absolute thermal energy of the cup and the water.
add the absolute energy of the cup to the absolute energy of the water to get a total energy
add the heat capacity of the cup and the water to get a combined heat capacity

divide the total energy by the combined heat capacity to get the equilibrium temp in Kelvin

convert temperature back to Celsius

NB 1 doesn't work unless you're using Kelvin for the calculations
NB 2 the heat capacity of the water is the SHC multiplied by the mass of the water (hopefully obvious)
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MSB47
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(Original post by Joinedup)
Delta T is a difference in temperature not the equilibrium temperature.

What's going to happen here is that the cup is going to increase in temperature and the water is going to decrease in temperature until cup and water are at the same temperature... Teqm

...and we're going to obey conservation of energy.

ΔTcup=Teqm-20 (we can see that it's going to be positive if Teqm is greater than 20)

ΔTwater=Teqm-90 (which is going to be negative if Teqm is less than 90)

so we can write

ΔQcup = C ΔTcup
ΔQwater=m c ΔTwater

ΔQwater=-ΔQcup (the reduction of the energy in the water equals the increase in energy of the cup)

and solve this


-----
alternative method (my 'absolute' favourite)

convert temperatures to Kelvin (add 273.15)
multiply the starting temp of the cup and the water by the appropriate heat capacity to get the absolute thermal energy of the cup and the water.
add the absolute energy of the cup to the absolute energy of the water to get a total energy
add the heat capacity of the cup and the water to get a combined heat capacity

divide the total energy by the combined heat capacity to get the equilibrium temp in Kelvin

convert temperature back to Celsius

NB 1 doesn't work unless you're using Kelvin for the calculations
NB 2 the heat capacity of the water is the SHC multiplied by the mass of the water (hopefully obvious)
I tried the second method I got the correct answer however for the first method im still confused because I dont know how to use the algebra :/ and for the second method why do you divide the total energy by the total heat capacity? either way thanks for your help its helped me alot

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Joinedup
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(Original post by MSB47)
I tried the second method I got the correct answer however for the first method im still confused because I dont know how to use the algebra :/ and for the second method why do you divide the total energy by the total heat capacity? either way thanks for your help its helped me alot

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in method 2 you're finding the temperature you get when you share an amount of thermal energy between the total heat capacity of the two objects combined.

you combine the heat capacities by addition, like you would if you had a question asking you to work out how long you'd need to microwave a cup of water in a microwave of a given power to raise the temperature of the cup and water (in thermal equilibrium) by 10 celsius

and you just share the energy by division.
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MSB47
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(Original post by Joinedup)
in method 2 you're finding the temperature you get when you share an amount of thermal energy between the total heat capacity of the two objects combined.

you combine the heat capacities by addition, like you would if you had a question asking you to work out how long you'd need to microwave a cup of water in a microwave of a given power to raise the temperature of the cup and water (in thermal equilibrium) by 10 celsius

and you just share the energy by division.
ohh that makes sense now, thanks for you help very much appreciated
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