# Thermodynamics: work done in a thermally insulated system

• randomgamernerd
In summary, the system described is a closed, thermally insulated copper vessel containing 200 g of water and weighing 100 g. After shaking vigorously for 15 minutes, the temperature of the water rises from 15°C to 17°C. The specific heat capacities of copper and water are given as 20 J/kg K and 4200 J/kg K, respectively. The questions posed are (a) how much heat is transferred to the liquid-vessel system, (b) how much work has been done on the system, and (c) how much is the increase in internal energy of the system? It is determined that no heat is transferred to the system, and the increase in temperature is a result of work done on the
randomgamernerd

## Homework Statement

: [/B]A thermally insulated closed copper vessel contains water at 15°C win the vessel is shaking vigourously for 15 minutes the temperature rises to 17°C the mass of the vessel is hundred gram and that the water is 200 g capacities of copper and water for 20 J/kg K and 4200 J/Kg K respectively . Neglect any thermal expansion (a)how much heat is transferred to the liquid -vessel system (b) how much work has been done on the system (c) how much is the increase in internal energy of the system?

## Homework Equations

: ΔU = Q - W
Q= m.s.ΔT...(1)[/B]

## The Attempt at a Solution

:[/B]
Okay, the first part:
Since the system is thermally insulated, heat cannot enter the system. So no heat is transferred.
Part B:
Q=(mwsw + mcsc)ΔT
= 1764 J
Now, the answer apparently matches with the text. My problem is I can't feel the "physics" here..I just blindly plugged the values into the formula..and ta da...I get the correct answer.
My question here is why did the temperature rise? and what exactly is the significance of equation 1.. I mean what does Q in the formula actually represent?? Is it amount of heat flowing in or out of a system?? No doubtedly some work is done here and that causes release of heat..The system is insulated..so the heat can't escape..so the temp rises..great..who does this work and how?

Part c:

Assuming that my part B is correct:
ΔU = Q - W
work is done on the system.
so ΔU = Q - (-1764)
= 1764 J [Q=O]

Maybe my calculations were wrong, but I got 1684 J for part B, based on the information provided in the problem statement.

I have wondered about this previously, but I don't have a good understanding of it. But what I believe is happening is that the "vigorous shaking" the friction in the water and also between the water and the copper vessel produces heat, causing both the water and the copper to heat up. (Maybe one of the Physics Forums experts will chime in and shed more light on this.)

Q represents heat (or energy). One way of looking at equation 1 is that Q represents how much energy is required for a certain amount (mass) of a substance to be increased by a certain temperature delta. So for water, it would take 4200 J of energy to increase 1 kg of water by 1 C°. Or that same 4200 J would increase 0.1 kg of water by 10 C°.

The unique aspect about this problem I see here is that water has such a large heat capacity - as you compare it with copper, for example. If, somehow, that same experiment was able to be performed with only copper (instead of copper and water), the increase in temperature would be more noticeable. Let's say that the 1684 J was imparted to 0.3 kg of copper, instead of 0.1 kg of copper and 0.2 kg of water. Then the temperature increase would be:
ΔT = Q/ms = 1684/(0.3)(376) = 14.9 C°. That is because water has such a large specific heat capacity as compared to copper. NOTE: I looked up the specific heat of copper and used that value (376 J/kg-K), instead of the 20 J/kg-K that was provided in the problem statement.

One time I heard somewhere that because of water's large heat capacity, you could dip your finger in water and then immediately dip it in a deep fryer of hot oil - in and out very quickly - and not get burned. (DO NOT TRY THIS AT HOME.) Of course, I had to try it. I just did it in and out very quickly and did not get burned. This was many years ago so I don't remember it very well. I remember I could feel heat, but it certainly did not burn me. I did not do a dry finger experiment as a comparison. :)

randomgamernerd
You are correct that Q = 0 for this, and that the temperature rise is the result of the work used to shake the vessel, which causes viscous dissipation of mechanical energy to internal energy of the contents plus the vessel. The error in all this is your 2nd Relevant Equation. In freshman physics, they incorrectly taught us that ##Q = mC\Delta T##. This equation is correct only if no work is done. In thermodynamics, we learn that correct equation for a nearly incompressible liquid should be:$$\Delta U=mC\Delta T$$ This is because, in thermodynamics, Q is a function of path (and many paths will give the same ##\Delta T##), while U and C are properly state functions.

TomHart said:
Maybe my calculations were wrong, but I got 1684 J for part B, based on the information provided in the problem statement.

I have wondered about this previously, but I don't have a good understanding of it. But what I believe is happening is that the "vigorous shaking" the friction in the water and also between the water and the copper vessel produces heat, causing both the water and the copper to heat up. (Maybe one of the Physics Forums experts will chime in and shed more light on this.)

Q represents heat (or energy). One way of looking at equation 1 is that Q represents how much energy is required for a certain amount (mass) of a substance to be increased by a certain temperature delta. So for water, it would take 4200 J of energy to increase 1 kg of water by 1 C°. Or that same 4200 J would increase 0.1 kg of water by 10 C°.

The unique aspect about this problem I see here is that water has such a large heat capacity - as you compare it with copper, for example. If, somehow, that same experiment was able to be performed with only copper (instead of copper and water), the increase in temperature would be more noticeable. Let's say that the 1684 J was imparted to 0.3 kg of copper, instead of 0.1 kg of copper and 0.2 kg of water. Then the temperature increase would be:
ΔT = Q/ms = 1684/(0.3)(376) = 14.9 C°. That is because water has such a large specific heat capacity as compared to copper. NOTE: I looked up the specific heat of copper and used that value (376 J/kg-K), instead of the 20 J/kg-K that was provided in the problem statement.

One time I heard somewhere that because of water's large heat capacity, you could dip your finger in water and then immediately dip it in a deep fryer of hot oil - in and out very quickly - and not get burned. (DO NOT TRY THIS AT HOME.) Of course, I had to try it. I just did it in and out very quickly and did not get burned. This was many years ago so I don't remember it very well. I remember I could feel heat, but it certainly did not burn me. I did not do a dry finger experiment as a comparison. :)
okay, thanks man..once again you helped me out..
And I will try this at home :p

TomHart
Chestermiller said:
You are correct that Q = 0 for this, and that the temperature rise is the result of the work used to shake the vessel, which causes viscous dissipation of mechanical energy to internal energy of the contents plus the vessel. The error in all this is your 2nd Relevant Equation. In freshman physics, they incorrectly taught us that ##Q = mC\Delta T##. This equation is correct only if no work is done. In thermodynamics, we learn that correct equation for a nearly incompressible liquid should be:$$\Delta U=mC\Delta T$$ This is because, in thermodynamics, Q is a function of path (and many paths will give the same ##\Delta T##), while U and C are properly state functions.
Okay..but what I've learned is
ΔU =nCvΔT
where Cv is molar heat capacity at constant volume..

In the second equation I referred to, s stands for specific heat capacity.
And I've been taught both are different...

randomgamernerd said:
Okay..but what I've learned is
ΔU =nCvΔT
where Cv is molar heat capacity at constant volume..

In the second equation I referred to, s stands for specific heat capacity.
And I've been taught both are different...
I left out the v subscript. In the 2nd equation, I've never seen an s used, but, yes, it's specific heat. The critical thing to note in this problem is that Q=0; and that mc delta T determines delta U, and not Q.

## 1. What is the definition of work done in a thermally insulated system?

The work done in a thermally insulated system is the energy transfer that occurs due to differences in temperature. This can be either in the form of work done on the system or work done by the system.

## 2. How is work calculated in a thermally insulated system?

The work done in a thermally insulated system is calculated using the formula W = PΔV, where W is the work done, P is the pressure, and ΔV is the change in volume of the system.

## 3. What is the significance of a thermally insulated system?

A thermally insulated system is important because it allows us to study the behavior of a system without any external heat or energy transfer. This helps us understand thermodynamic processes and make accurate predictions about the behavior of the system.

## 4. Can work be done in a thermally insulated system without any change in temperature?

No, work cannot be done in a thermally insulated system without any change in temperature. This is because work is a form of energy transfer, and without a change in temperature, there is no difference in energy to cause the transfer.

## 5. How does the first law of thermodynamics apply to work done in a thermally insulated system?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted from one form to another. In a thermally insulated system, the work done is a form of energy transfer, and therefore, the first law of thermodynamics applies to it.

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