Thermodynamic ( quanity of heat)

[Newton86]

[SOLVED] Thermodynamic ( quantity of heat)

Homework Statement

Calculate he quantity of heat that needed to transfer 2,0 kg ice of -10C to steam/vapor of 100C
The process in this case take place in a good heat-isolated container with heat capacity of 500J/Kg

Homework Equations

Q=mcT
Q+-ml ?

The Attempt at a Solution

Hi this is what i done
I find that it takes 225600j to change 1 kg liquid water from 100C to vapor.To find the required heat for the temprature change
for the water to change from 100C to vapor: 225600J * 2kg = 451200J

For the ice to melt and heat to 100C: 2kg*2100j*110= 462000J
!: Not sure if this is the right way to find the ice heat since its a phase change it is done diffrently ?

adding these two finds the heat needed to transfer the ice to vapor ?

But what do i do about the heat capasity I am not sure what to do. It does have a effect but i don't know how.

 

[Shooting Star]

Break it up into five parts:

1. Ice heats up to melting pt.
2. Ice melts.
3. Water heats up to boiling pt.
4. Water boils.
5. Container heats up from initial temp to final temp.

Find the quantity of heat required for each part and add. Use the appropriate eqns for each part.

 

[Redbelly98]

One must know the mass of the container to do this. (Item #5 in Shooting Star's list)

 

[Newton86]

The container is not inportnant in this task it says its uninportant. So it must be solved another way then.

 

[Redbelly98]

It's still solved the same way, except that you can ignore Step #5.

 

[Newton86]

ok I will try to solve it this way then.

 

[Newton86]

Hmm =| Off course the container is inportnant. I mixed up. It has a heat capacity of 500J/Kg
But it doesn't have a mass. So what I do with it ?

 

[Redbelly98]

Hmm =| Off course the container is inportnant. I mixed up. It has a heat capacity of 500J/Kg
But it doesn't have a mass. So what I do with it ?

That's puzzling. But first, I am confused about the units in the figure you give. Does "Kg" mean "kilograms" or "Kelvins * grams"? Please double-check the units given in the problem statement.

You will need the container mass, since it absorbs heat energy of

$$\Delta E = c \cdot M \cdot \Delta T$$

Mass M is needed for the calculation.

 

[Kaleb]

Also set your equations equal to one another. Energy lost = energy gained ect...I had the same problem awhile back and that's what I was leaving out.

 

[Newton86]

That's puzzling. But first, I am confused about the units in the figure you give. Does "Kg" mean "kilograms" or "Kelvins * grams"? Please double-check the units given in the problem statement.

You will need the container mass, since it absorbs heat energy of

$$\Delta E = c \cdot M \cdot \Delta T$$

Mass M is needed for the calculation.

Aii yes it is 500J/K not Kg :)

I tryed to find the heat gained from ice to vapor.

1: Ice heats up to melting point
Q=mcdeltaT = 2kg*2100j/kg*K*10K = 42000J


2: Ice melts at C0
Q=+-ml = 2kg* 334000J = 668000J

3: liquid water of C0 heats up to 100C
Q=mcdeltaT = 2kg*4190j/kg*K*100K = 838000J

4: water of 100C boilis up to vapor
Q = +-ml = 2kg*2,256*10^6 = 225600J

All 4 parts added = 3804000J

this correct ?
But now it the problem I don't know what to do with the heat capasity.

 

[Shooting Star]

(Repost)

Also set your equations equal to one another. Energy lost = energy gained ect...I had the same problem awhile back and that's what I was leaving out.

What is losing energy? Have you gone through the problem?
--------------------------------------------------------

(Before Newton86's last post.)

It's possible that the thermal capacity of the container is given, the unit of which is J/°K, and not the specific heat. Then the mass is not necessary. Only the OP can give us the proper info.
--------------------------------------------------------

OK, so you have found the proper units.

Thermal capacity = m*c, which is the heat required to raise the temperature of the object by 1°K. The specific heat c is the heat required to raise the temperature of unit mass of a substance by 1°K.

In that case, heat gained by object = 500*∆t.

 

[Redbelly98]

Shooting Star is right, the 500 J/K is $$c \cdot M$$. They have already combined c and M into one value.

 

[Newton86]

hmm. So what I do is to find the temprature out of thoose 4 parts and * it with the 500J/K ?

 

[Redbelly98]

If I understand you correctly, yes. You find the temperature change for those 4 parts.

 

[Newton86]

Ok I think what I don't seems to find now is the

Also set your equations equal to one another. Energy lost = energy gained ect...I had the same problem awhile back and that's what I was leaving out.

hmm is the energy lost in the phase changes ?

 

[Redbelly98]

If I understand you correctly, yes. You find the temperature change for those 4 parts.

I think I didn't say that very well.

By how much does the temperature of the container change? I.e., what are the starting and ending temperatures of the container?

You're almost there. You have done 4 out of the 5 calculations, you just need to complete #5 and add up the energies.

 

[Newton86]

I think I didn't say that very well.

By how much does the temperature of the container change? I.e., what are the starting and ending temperatures of the container?

You're almost there. You have done 4 out of the 5 calculations, you just need to complete #5 and add up the energies.

hmm Starting temprature of the container must be -10 Celsius and ending temprature of 100Celsius ?

So then the heat from container is 500j/k *110K = 55000J ?

 

[Redbelly98]

Yes, 55000 J of heat is required to heat the container.

 

[Shooting Star]

2: Ice melts at C0
Q=+-ml = 2kg* 334000J = 668000J

For consistency, you should use the proper units in this step and step 4 too, and write the latent heat of fusion with units of J/kg. Don't use +-; since you are giving heat here, use the +.

In general, if you are using a consistent set of units, stop putting the units inside the eqn -- it just looks messy.

 

[Redbelly98]

Personally, I prefer keeping the units within the calculation. It's provides a check on:

1. Did I forget any terms? The answer will usually have the wrong units if I did.

2. Did I forget to convert any of the units? I often see problems where m and cm are used to specify different lengths. If the answer contains something like (m/cm) in it, I can simply replace that with a 100 to get the final answer.

But, that's another topic and perhaps best left for another thread.

 

[Newton86]

For consistency, you should use the proper units in this step and step 4 too, and write the latent heat of fusion with units of J/kg. Don't use +-; since you are giving heat here, use the +.

In general, if you are using a consistent set of units, stop putting the units inside the eqn -- it just looks messy.

Ok :) thanks for the advice.

But what do I now
Is it just to add the heat from the container 55000J too the 4 parts added = 3804000J
= 3804000J + 55000J = 3859000J ?

 

[Redbelly98]

Yes.

 

[Newton86]

:D so that's the answer then 3859000J
Thanks for the help :)

 

[Shooting Star]

Personally, I prefer keeping the units within the calculation. It's provides a check on:

1. Did I forget any terms? The answer will usually have the wrong units if I did.

2. Did I forget to convert any of the units? I often see problems where m and cm are used to specify different lengths. If the answer contains something like (m/cm) in it, I can simply replace that with a 100 to get the final answer.

But, that's another topic and perhaps best left for another thread.

I, too, don't want to continue this discussion here, but I can't resist mentioning a couple of useful points here. If you are very good in Physics and Math, then for you the way you said may not pose a problem.

Ideally, the mathematics should be done using symbols, and the values with the units should be substituted only at the last step. I have seen students put a known big value at the first step and it's dragged along throughout the calculation maybe for 12 steps with the other symbols, sometimes 4 becoming 3 somewhere on the way. So, numerals should be avoided until the last possible moment.

I mentioned using a consistent system of units, which implies that the m has been converted to cm or vice versa beforehand. For a fresher, dealing with m and cm in the same eqn will be disastrous.

And lastly, you obviously haven't encountered somebody who canceled mass m in the denominator with the m for meter in the numerator. I have! Now you understand my point, I hope.

Enough said.

 

[Redbelly98]

Understood.