Thermal contact final temprature

In summary, the equation used in the given problem involves the heat capacities of two masses, which can be solved for the final temperature by equating the initial and final heat. The concept of heat capacitance, denoted as C, is used to simplify the equation by incorporating the mass and specific heat capacity of an object into one value. This is a common practice in thermodynamics to make writing equations easier.
  • #1
roam
1,271
12

Homework Statement



The following is a worked problem from my textbook:

Consider two masses, with initial tempratures T1 and T2 are placed in contact and isolated from their surroundings. They are

CV1T1 + CV2T2 = CV1Tf+CV2Tf

This equation can be solved for the final temprature

[itex]T_f = \frac{C_{V1}}{C_{V1}+C_{V2}} T_1 + \frac{C_{V2}}{C_{V1}+C_{V2}} T_2[/itex]

So, if I understand it correctly they have used the idea that the total internal energy is conserved, and they equated the initial to the final Q. Is that right? I'm a bit confused as how they got the expression "CvT"? But shouldn't they be using Q=mcvΔT? Why did they omit "m"?

The Attempt at a Solution



I can use Q=mcvΔT and equate heat lost by the hotter mass to the heat gained by the colder mass. We don't have the mass, but I don't think we can just omit that. So, how did they get "CvT"? I'm a bit confused. Any help is greatly appreciated.
 
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  • #2
roam said:

Homework Statement



The following is a worked problem from my textbook:

Consider two masses, with initial tempratures T1 and T2 are placed in contact and isolated from their surroundings. They are

CV1T1 + CV2T2 = CV1Tf+CV2Tf

This equation can be solved for the final temprature

[itex]T_f = \frac{C_{V1}}{C_{V1}+C_{V2}} T_1 + \frac{C_{V2}}{C_{V1}+C_{V2}} T_2[/itex]

So, if I understand it correctly they have used the idea that the total internal energy is conserved, and they equated the initial to the final Q. Is that right? I'm a bit confused as how they got the expression "CvT"? But shouldn't they be using Q=mcvΔT? Why did they omit "m"?


The Attempt at a Solution



I can use Q=mcvΔT and equate heat lost by the hotter mass to the heat gained by the colder mass. We don't have the mass, but I don't think we can just omit that. So, how did they get "CvT"? I'm a bit confused. Any help is greatly appreciated.

It's a bit like why in electronics we rate capacitors with a given capacitance rather than specify their plate area, dielectric constant, and plate separation separately; a lumped constant representing the total charge versus voltage is more convenient.

For thermal "circuits" it can be convenient to take Q=mcΔT for each fixed object or substance and turn it into Q = CT, with T now the absolute temperature and C = mc the 'heat capacitance' for the given object. It can make writing the equations for the system a bit easier if you don't have to keep dragging around individual masses and ΔT's and so forth.

The heat capacitance for a given object contains the same information as if you were given its mass and heat capacity coefficient and took the trouble to multiply them together everywhere you needed to insert them in an equation :wink:
 
  • #3
It's common in thermodynamics to write specific (intrinsic) quantities in lowercase and absolute (extrinsic) quantities in upper case. These kinds of measurements are referred to as intrinsic and extrinsic properties, respectively, where an intrinsic property is independent of the amount of "stuff" present (density, temperature, etc) and an extrinsic property depends on the amount of "stuff" present (weight, volume, etc).

For example, specific volume is volume per unit mass and portrayed with a lowercase v, and has the units of m[itex]^{3}[/itex]/kg, while absolute volume is portrayed with an uppercase V, and has the units of m[itex]^{3}[/itex]. In your case, the specific heat capacity is written as c[itex]_{v}[/itex] and has the units of J/(kg[itex]\cdot[/itex]K), while the heat capacity is written as C[itex]_{v}[/itex] and has the units of J/K.
 
  • #4
Oh, I get it now. The small letter is the heat capacitance per unit mass so that's why the masses cancel out, and it's more convinient not to write it. Thank you very much for the help! It makes perfect sense now. :)
 
  • #5
Glad I could help!
 

1. How is thermal contact final temperature calculated?

The thermal contact final temperature is calculated using the equation Tfinal = ((m1 * c1 * T1) + (m2 * c2 * T2)) / (m1 * c1 + m2 * c2), where m1 and m2 are the masses of the two objects, c1 and c2 are their specific heat capacities, and T1 and T2 are their initial temperatures.

2. What factors affect the thermal contact final temperature?

The thermal contact final temperature is affected by the masses and specific heat capacities of the objects in contact, as well as their initial temperatures and the duration of the thermal contact. Other factors such as thermal conductivity and the presence of insulation also play a role.

3. How does thermal equilibrium relate to thermal contact final temperature?

Thermal equilibrium is the state where two objects in contact reach the same temperature. In the case of thermal contact final temperature, the objects reach thermal equilibrium when their final temperatures are equal.

4. Can the thermal contact final temperature be higher than the initial temperature of one of the objects?

Yes, it is possible for the thermal contact final temperature to be higher than the initial temperature of one of the objects. This can happen if the other object has a higher initial temperature and a larger mass or specific heat capacity.

5. How does the thermal contact final temperature differ from the final temperature of a single object?

The thermal contact final temperature takes into account the properties of two objects in contact, while the final temperature of a single object only considers its own properties. Thus, the thermal contact final temperature may be different from the final temperature of a single object after being heated or cooled.

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