# What is the concept behind these two similar Thermodynamics problems

1. May 22, 2014

### hbk69

1. The problem statement, all variables and given/known data

Question 1) a mass m=200g or an unknown metal is heated to a temperature T1=200°C and then dropped into m=50g of water at a temperature T2=20°C in an insulated container. The water temperature rises within a few seconds to T3=39.7°C and then changes no further. Specific heat capacity of water=4190 J/kgK. Find the specific heat capacity of the metal cM.

Question 2) A small metal container holds water of a mass m=20g at a temperature T=20°C. Another large ridig container with Volume V=4000cm^3 holds n=0.4mol moles of monoatomoic gas at a pressure p1=10^6pa and a temperature T1. The metal container and the large container are placed in good thermal contact with each other. Both containers are perfectly insulated from the outside. The containers are left like this for a long time so thermal equilibirum is established. Specific heal of water is c=4186 J/Kg*K. Find the pressure p2 after the containers are in thermal equilibirum.

2. Relevant equations

Q=m*c*ΔT
pV=n*R*T
ΔEth=-W+Q
ΔEth=3/2*n*R*ΔT
p/T=constant

3. The attempt at a solution

I have the solutions for both problems but there is an aspect of the solutions which my understanding needs a great deal of improvement.

Question 1)

In this problem the heat energy in water (Qw) gains heat energy because when the hot metal is dropped in the water its heat is transffered to the water, so the hot metal loses energy (Qm)?

There for Qw = -Qm

Question 2)

In this problem there is a temperature difference between the water container and the gass container so we know there will be heat transfer. Since the calculated Temperature of the gas T1 is greater then the temperature of water T=20 it is said that the heat Q flows from the container to the water container so in other words if one object has greater temperature then the other heat will transfer from the high temp object to the low temp object?

Since heat will transfer from the gas container to the water, you'd think the water gains the energy and the gas loses the energy. So the heat energy of water (Qw) = -Q(heat energy of the gas)

But in the solution this is not the case and the heat energy of the gas Q = -Qw implying that the gas gains energy while the water loses energy.

Where am i going wrong in the questions? the prolems i have in them are very similar, i believed i have not read the scenarios correctly.

Any help great appreciated, thanks guys.

2. May 22, 2014

### Staff: Mentor

While this may imply the water loses energy, it may be unreasonable to interpret it that way. That equation applies regardless of Qw being positive or negative. If the data indicates otherwise, then go with the data. (You calculated in Kelvin temperatures?)

Where heating of liquid is involved, always allow the possibility that it may change state.

3. May 22, 2014

### hbk69

In the question 2 problem i understand that the heat Q of the gass flows from the gas to the water but don't understand how the Qw energy of the water becomes negative because it is gaining energy and in question 1 when the Qw of water gains energy from the metal it remains positive hence the confusion i have with the problem in question 2.

Maybe in question 1) Qw = -Qm but the water does not gain energy and the metal does not lose it? instead its the opposite

and in question 2) since the water is gaining the energy it will be -Qw while the Q of the gas remains positive as it is providing the energy in the first place.

But i can't get my head around that i don't understand how something can be positive if it is losing energy while transffering heat to the water in question 2 for example

4. May 22, 2014

### dauto

(Qw) = -Q
and
Q = -Qw
are the same equation written in two different ways, so I have no idea what you're complaining about.

5. May 22, 2014

### Staff: Mentor

A negative heat gain is the same as a positive heat loss.

Perhaps it's time to show us your working.

6. May 23, 2014

### hbk69

Question 1) Qw(heat energy of water) = -Qm(heat energy of metal)

In this scenario why is the Qm negative? is it because the metal loses heat while the water gains heat energy?

Qw=c(specific heat capacity of water)*mw(mass of water)*(T3-T2)

Qm=cm(specific heat capacity of the metal)*m(mass of metal)*(T3-T1)

as Qw = -Qm

then c(specific heat capacity of water)*mw(mass of water)*(T3-T2)= -cm(specific heat capacity of the metal)*m(mass of metal)*(T3-T1)

then to we rearrange to find cm the specific heat capacit of the metal:

cm=((c*mw*(T3-T1)/m*(T1-T3))

Question 2)

Initial temperature of the gas T1:

p1*V=n*R*T1

T1=p1*V/n*R

T1=1203k which is greater then T=20°C of the water therefore the heat flows from the gas container to the water

Qw= is the th heat energy that enters the water container.

Qw=m(mass of water)*c(specifict heat capacity of water)*(T2-20°C)

To find the state of the gas:

ΔEth=-W+Q

No work is done so ΔEth=Q for the gas

Therefor Q of the gas = -Qw of the water

Q(gas)=-Qw(water

Why is the -Qw negative? i know the heat energy is transffered from the gas to the water container but shouldn't heat be lost from the gas and gained by the water? so -Q=Qw because even in question 1 the water gains energy while the metal loses energy.

I have the correct solution but as you can see i don't understand why Qw = -Qm in question 1 and why Q=-Qw for question 2 because i feel that for question 2 it should be -Q=Qw given the reasons stated. This was the reason why i did not post the full working out in the OP felt i would not need to.

i will still continue the solution to question 2 since you had asked:

As Qw=m(mass of water)*c(specifict heat capacity of water)*(T2-20°C) and ΔEth=n*R*(T2-T1)

n*R*(T2-T1)= - m(mass of water)*c(specifict heat capacity of water)*(T2-20°C)

This is rearranged to find T2 which is calculated to be 72°C

Then i use Gas-Lussac Law p/T=constant to find the pressure p2.

The solutions are fine they are from worked examples but i remain confused over the energies and how to anticipate which need to be postive or negative when you have to e.g Qw=-Qm etc

Any help much appreciated

7. May 23, 2014

### hbk69

So you are saying it does not matter which one you put as the negative, there will always be one positive Q and one negative Q, but that could possibly cause some problems when you are rearranging the formula when you are trying to find a specific variable

For example in question 2

Q=-Qw

n*R*(T2-T1)= - m(mass of water)*c(specifict heat capacity of water)*(T2-20°C)

when i rearrange the formula in-order to find T2 i get T2=3/2*R*T1*m*T/3/2*n*R+m*c, will i still get the same if i have -Q=Qw?

thanks

8. May 23, 2014

### D H

Staff Emeritus
Yes.

However, the way you asked this question indicates a misunderstanding: It appears you think objects contain heat. Just as work is something that can be performed on or by the object, heat is something that can be transferred to or from the object. However, the object does not contain work or heat.

Heat and work together collectively act to change the energy of the object, which is something the object does contain. The relationship between heat, work, and internal energy is the subject of the first law of thermodynamics.

Because heat and work are things transferred rather than contained, there's a bit of ambiguity with regard to the sign convention. With heat it just makes sense to define heat as positive if it makes the object gain energy, negative if it makes the object lose energy. With work, it's not so obvious. Some define work as positive if the object performs work on the environment, others define it exactly the opposite. The first law of thermodynamics is $\Delta U = Q-W$ with the first convention but is $\Delta U = Q+W$ with the latter convention.

Getting back to heat, which direction is positive and which is negative is just a convention. It's nice to have a universally consistent convention. Compare with work, which isn't consistent across the literature. This makes for a pain in the rear when reading different authors. That said, everything would work out just fine if heat flow had been defined to be positive when the object is transferring heat to the external environment. The equations would just have different signs.

With regard to these two problems, both are specifically set up so that work is zero, and both are specifically set up so that the only thing you have to worry about thermally are the two objects in each question. The water+metal system in question #1 and the water+gas system in question #2 are both isolated from the external environment. All you have to worry about is heat transfer.

9. May 25, 2014

### hbk69

In the given problems i don't understand the reasoning behind the convection which is used and that would indicate i don't understand what is going on in the problem, with regards to question 1 my understanding is clear now given that you confirmed that the metal loses heat while the water gains heat energy hence Qw(heat energy of water) = -Qm(heat energy of metal).

But for question 2)

Q(gas)=-Qw(water

Why is the -Qw negative? i know the heat energy is transffered from the gas to the water container but shouldn't heat be lost from the gas and gained by the water? so -Q=Qw because even in question 1 the water gains energy while the metal loses energy.

10. May 28, 2014

### hbk69

Bump, any help for the problem i have in the above post. Have an Exam soon Thanks for any guidance

11. May 28, 2014

### D H

Staff Emeritus
There's an implicit assumption with question #2, which is that both the rigid container that contains the gas and the bowl that contains the water have negligible heat capacity. With this assumption, the only things that contribute to the equilibrium state are the gas inside that rigid container and the water inside that bowl.

12. May 28, 2014

### TheAustrian

I would like to add, that from my experiences, $\Delta U = Q-W$ is a more commonly used convention.

13. May 28, 2014

### D H

Staff Emeritus
I agree, but it is not universal. It's not that hard to find a textbook or article that uses $\Delta U = Q+W$.

That Q denotes heat flow into a system (as opposed to heat flow out of a system) is, as far as I can tell, a universal convention.

14. May 28, 2014

### sankalpmittal

In physics we use ΔU=Q-W with signs.

Work done by system is positive and work done on system is negative.

However in chemistry,

ΔU=Q+W

Work done by system is negative and work done on system is positive.

Just difference in conventions.

15. May 28, 2014

### George Jones

Staff Emeritus
These days, the $\Delta U = Q + W$ convention is used quite often in physics texts. For example, the text "Thermal Physics" by Schroeder is widely used in North America and it uses this convention. This convention also is used in the first-year physics by Serway and by Knight. Over the last couple of years, I have used all of these as texts for physics courses that I have taught.

The first-year book by Halliday, Resniick, and Walker, however, uses the other convention. So, both conventions are used extensively, and students should be careful.

At my university students see thermal/statistical physics in first-year in either Knight or Serway (+W), in second-year in Schroeder (+W), and in fourth-year in Baierlein (-W).

16. May 29, 2014

### TheAustrian

+1. This post completely in accordance with my observations. I've only seen ΔU = Q - W in Physics, exclusively. Thermo is a subject that props up in both Physics and Chemistry, but it's used in a different manner in each discipline. Also, when it comes to writing thermodyanamic potentials, ΔU = Q - W seems to make more sense to apply.