# Homework Help: Simple concept about Q=mcΔT

1. May 14, 2012

### hqjb

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

I'm confused by the different equations textbooks use for Q=mcΔT

I understand for solids/liquids its always Q=mcΔT but for gases am I right to say ΔH=Q=ΔU+pΔV=m(c_p)ΔT for constant pressure and Q=ΔU=m(c_v)ΔT for constant volume?

So if my textbook says "ΔU=n(3/2R)ΔT for any processes of monoatomic ideal gas" its wrong because for constant pressure processes you need to add work? Or am i getting something wrong.

Edit:

What I really want to know is if I should use Q=nCt instead of U=nCt[\b] as the book did and why not(I think the book should be right...)

Last edited: May 14, 2012
2. May 14, 2012

### truesearch

Strictly speaking you should always start with ΔQ = ΔU + ΔW.
It applies to solids and liquids as well as gases.
ΔW (external work) is only zero if there is no expansion. For a gas this means keeping the volume constant.
For solids and liquids any expansion is usually negligible and it is safe to ignore any energy needed to do work ΔW against the atmosphere.
It is not too difficult to look up coefficients of expansion to get an estimate of how much energy is needed to do external work ΔW
It means that for gases there are 2 principal specific heats. SHC at constant pressure and SHC at constant volume. The SHC at constant pressure is greater than the SHC at constant volume

3. May 14, 2012

### Infinitum

No. The specific heat you are taking into consideration while calculating for gases is molar specific heat at constant pressure and volume respectively. So there would be a term of number of moles (n) rather than mass (m).

No, the textbook is correct. Can you try explain that based on the correction above?

4. May 14, 2012

### hqjb

a)So for gases theres no specific heat only molar specific heats?

b)But according to the post by truesearch for gases work done is not negligible and
Q=U+W=nCt but in the question the gas expands so work done should not be negligible why did the answer use U=nCt.
I understand 3/2R is the approximation for SHC at constant volume(so shouldnt it not be used since volume is not constant?)

5. May 14, 2012

### truesearch

It is possible to switch between SHC and molar SHC quite easily.....does not affect the physical explanation.

6. May 14, 2012

### Infinitum

The specific heat of gases is generally measured as molar specific heat. You can still say they have specific heat, because by definition, specific heat is the is the amount of heat per unit mass required to raise the temperature by one degree Celsius. Its only convenient and more informative to use the molar specific heat, since it considers the two important factors for gases, i.e pressure and volume.

Which C is that? Cp or Cv? Then think of the relation between them...

7. May 14, 2012

### hqjb

Yes! I think I've got it.
Since Cp = Cv + R
if I use Cv which is Cp - R when I do Q=nCvt=n(Cp-R)t=nCpt - nRt, the nRt portion already takes into account work done? So its really not Q=nCvt but U=nCvt but for constant pressure Q=nCpt=U
Thanks both for helping btw.

8. May 14, 2012

### Infinitum

Uhh nope. For constant pressure, its $\Delta U + W = nC_p \Delta T = \Delta Q$

The change in internal energy still remains

$\Delta U = nC_v \Delta T$

The work is included in the constant pressure equation since $C_p$ is involved.

Another way to understand this :- http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/shegas.html#c1

The internal energy change is expressed by the kinetic theory of gases :

$\Delta U = nN_A\Delta KE_{avg} = \frac{3}{2}nN_Ak\Delta T = \frac{3}{2}nR\Delta T$

9. May 14, 2012

### hqjb

Sorry brainfart ignore that. I meant I could use Q=nCvt for constant pressure processes but in that case its not Q=nCvt but U=nCvt but if its for constant volume processes than its Q=nCvt=U.

10. May 14, 2012

### Infinitum

Yup! dU for an ideal gas is always equal to nCvdT