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# Thermodynamics Basics Calculation

thermodynamics 1st law energy balance chemical engineering calculate heat capacity

7 replies to this topic
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### #1 Em_And_Chem

Em_And_Chem

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Posted 11 September 2023 - 04:44 AM

I can't seem to get the right answer, any help is appreciated!
I have a question under the topic "First law of thermodynamics", I've also been told that I should use the complete Energy (or perhaps entropy?) balances.
The question goes as follows; A body of iron with constant heat capacity 460 J/(kg K) and mass 3 kg is heated with a heating effect of 500 W. How long does it take to heat the body from 250C to 1000C if the heat loss to the surroundings can be estimated at 15 W?

so far Ive (perhaps incorrectly) assumed:

The answer is apparently 213 seconds

Edited by Em_And_Chem, 11 September 2023 - 04:47 AM.

### #2 breizh

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Posted 11 September 2023 - 05:01 AM

Hi,

We don't do homework, but we are happy to help.

There is discrepancy between your problem statement and calculation.

25 C or 250 C ; 100 C or 1000 C ?

Probably good to review the units of the first equation .

Good luck

Breizh

### #3 Em_And_Chem

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Posted 11 September 2023 - 05:34 AM

Thanks for your response! Its 100 C. I did finally get to the answer, except I dont quite see how it links to the energy or entropy balances... Might you have an incling into how they could be linked?

### #4 breizh

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Posted 11 September 2023 - 06:28 AM

Hi,

Reading your first post I'm quite worried about your understanding of the fist law applied to open system, in particular your understanding of the terms in the first equation.

Underneath a few videos to support your work

Q=m*cp*(Tfinal -T start)= ( Power in - Power out (losses)) *time.

Good luck and enjoy the movies

Breizh

### #5 snickster

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Posted 11 September 2023 - 06:29 PM

Thanks for your response! Its 100 C. I did finally get to the answer, except I dont quite see how it links to the energy or entropy balances... Might you have an incling into how they could be linked?

You original equation is a first law energy balance steady flow equation.  Say you have a control volume such as a tank of liquid or whatever.  So on the left side you have the change in energy per unit time of the energy contained inside of the tank =

= Change in energy per unit time = dU/dt inside the tank (control volume)

The change in energy per unit time in the control volume is equal to the engergy per unit time added to the volume minus the energy removed per unit time.  This is equal to:

Heat (added or removed) per unit time = Qdot   plus+

Work (added or removed) per unit time = Wdot   plus+

Mass in minus mass out per unit time x the energy contained in that mass = Sum Mdot x (enthalpy/unit mass+KE per unit mass+PE per unit mass) for both in and out flow streams.

So your equation is one of a balance of energy based on the first law of thermodynamics.

But you don't have a flowing system.  Your control volume is the piece or iron itself with no mass flowing in or out.  I guess you could make it a flowing system say if you start with a 3kg piece of iron in a container then flow some molted iron in and out over a period of time and keep track of how much energy flows in versus out to find the net energy in or out of the system, but you don't have a flowing system - your system is the 3 kg of iron.

Therefore the terms involving mass in and mass out go away.

Now since you have a solid piece of iron that cannot change in volume you cannot have any work into or out of the system unless you drag the iron on the ground and cause frictional work to add energy to the iron in the form of heat but you don't have any movement of the iron either - so there is no work in or out of the system.

So that leaves you with the equation:

dU/dt = Qdot

Qdot is the heat added or removed from the system which in this case there is a net heat added of:

Qdot = (500 - 15) watts

So this is the increase in the internal energy of the system dU/dt

For any solid or liquid the increase in internal energy is (m)(Cp)(dT) since again there are no moving boundaries in the solid or liquid  then all energy input as heat goes in to iincreasing the interal energy (increasing molecular vibrational/kinetic energy) since there is no moving boundaries like when you add heat energy to a gas where some of the energy may go into internal energy (kinetic energy of molecules - temperature) and some may go into moving boundaries (such as piston work).  In the case of solid or liquid all heat added goes into kinietic/vibrational energy of molecules.

So the resulting equation is:

dU/dt=Qdot

= (Mdot)(Cp)(dT)/dt = 485 Watts.

Just like you solved in you last post above.

So the first law says that energy cannot be created or destroyed.  All the energy in the universe already exist someplace and in some form in the universe.  It can only be converted from one form into the other.  So you are like an accountant just keeping track of where all of it goes.

Edited by snickster, 11 September 2023 - 08:59 PM.

### #6 snickster

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Posted 12 September 2023 - 06:59 PM

Another way to look at it.

You have a flowing system of solid iron (3kg) flowing into a tank/container control volume.  Solid iron can't flow but just imagine it can.  Net heat is added to the iron 485 watts as it passes through the container control volume.  The iron then flows out of the container with all of the heat input leaving , all mass leaving that went in, and no mass or energy left in the container/control volume.  In this case:

Change in energy of the container/control volume = dU/dt = 0

Work = 0 as the iron passes throught the container since no work done on it

Therefore the energy equation reduces to:

Qdot = Mdot(Hin-Hout)      Since no change in kinetic energy or potential energy

dU/dt= M (Cp) (dT)   or

dt = dU/{(M) (Cp) (dT)}

Edited by snickster, 12 September 2023 - 07:02 PM.

### #7 Em_And_Chem

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Posted 15 September 2023 - 02:56 AM

Thank you very very much everyone!! Much appreciated.

### #8 snickster

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Posted 15 September 2023 - 09:55 PM

I see an error.  I should have wrote in my last post:

dU/dt= M (Cp) (dT) /dt  or

dt = dU/dt/{(M) (Cp) (dT)}   Where dU/dt = 485