0
$\begingroup$

In my physics book (and wikipedia) it states that the thermal energy transferred to an object is: $Q = c \ m \ dT$, where $Q$ is the transferred thermal energy, $c$ is the specific heat capcity of the material, $m$ is the object's mass and $dT$ is the change in temperature.

So far so good, but what I don't understand is that, as an objects temperature is changing, then surely its mass would change as well, right?

So how should one use the formula in that case? Is it as simple as the average mass between the two temperatures? Or should one just use the mass from the first state?

Improve this question
$\endgroup$
7
  • $\begingroup$ Why do you think the mass would change? $\endgroup$
    – mcFreid
    Jan 1, 2014 at 16:51
  • $\begingroup$ Partly because you always hear "warmer air rises because it's lighter" and partly because wolframalpha supports that thought. Air density at 10celsius: wolframalpha.com/input/?i=density+of+air+10celsius Air density at 20celsius: wolframalpha.com/input/?i=density+of+air+20celsius Is this not the case? $\endgroup$
    – Mickez
    Jan 1, 2014 at 16:54
  • $\begingroup$ I think you're confusing mass with density. Warmer air is less dense (i.e. less number of oxygen atoms per cubic meter) than cold air and rises because it is less dense. $\endgroup$
    – mcFreid
    Jan 1, 2014 at 16:57
  • $\begingroup$ Yes but isn't mass defined by density * volume? So if the density is lower and volume is the same, then surely the mass would be lower as well? $\endgroup$
    – Mickez
    Jan 1, 2014 at 16:58
  • $\begingroup$ If you fix the volume, then the density remains fixed regardless of temperature. $\endgroup$
    – mcFreid
    Jan 1, 2014 at 17:00

2 Answers 2

Reset to default
1
$\begingroup$

The mass does not change, the volume changes, which is why the density changes.

Considering the thing you're heating to be an ideal gas, the quantity (number of moles), volume, pressure and temperature are related as

$$ PV = nRT $$

P = pressure,
V = volume,
n = number of moles,
T = temperature
R = ideal gas constant = 8.314

When you're heating a gas in a closed container, the quantity (n) is constant, so there must be an increase in the pressure and/or volume to satisfy the equation.

Improve this answer
$\endgroup$
0
$\begingroup$

In thermodynamic analysis varying scenarios regrading pressure, volume, and temperature are imagined. The ideal gas law provides the unifying relationship (PV=nRT). If you were considering a fixed volume gas in contact with a solid mass there would be no change of volume with changes in temperature but the pressure would change and need to be taken into account. If you wanted the pressure to stay constant it would require a piston to maintain constant pressure and allow the volume to decrease as the temperature decreased.

If you were imagining a process with with constant pressure you need to consider the energy changes in the gas volume described for isobaric processes: http://en.wikipedia.org/wiki/Isobaric_process

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.