# Temperature, heat, and energy

1. Jul 12, 2011

### gkangelexa

Hey,
The internal energy E of a system has 2 components: kinetic energy and potential energy.

$\Delta$E = q + w

where w is work and q is heat. heat is the transfer of thermal energy
at constant pressures, q becomes H (enthalpy)

My book says that another way to calculate the internal energy change of a gaseous reaction is to assume ideal gas behavior and constant temperature. since PV = nRT

In which case the equation would be:

$\Delta$E = $\Delta$H - $\Delta$(PV)
$\Delta$E = $\Delta$H - $\Delta$(nRT)
$\Delta$E = $\Delta$H - RT$\Delta$n

My question is:
I know that heat is the transfer of thermal energy, and that it is a part of the total internal energy. And thermal energy depends on the kinetic energy of the molecules. But how does temperature relate to all this?

How can you have a change in heat (enthalpy) and no change in temperature?
(as the above equation shows)

Last edited: Jul 12, 2011
2. Jul 12, 2011

### Studiot

A simple example would be change of state (eg melting)

edit I should add 'of a pure substance.'

3. Jul 12, 2011

### gkangelexa

In the context of thermodynamics, kinetic energy is also referred to as thermal energy.
and the average kinetic energy of the molecules is proportional to the temperature of the gas.

Knowing this, then it makes sense that adding heat (thermal energy) should increase the temperature of the gas because you are adding kinetic energy
and temperature is proportional to the average kinetic energy...

so why doesn't it always work like that?

what is wrong in my thinking?

4. Jul 12, 2011

### Studiot

What does melting ice have to do with gas?

By the way it is not a good idea to post different questions with the same title in different forums. It is very confusing.

I think the author of the article Berkeman referred to depracted the use of the term thermal energy as confusing as well?

5. Jul 12, 2011

### gkangelexa

sorry yeah youre right the titles are confusing :-(

So is my thinking correct?

6. Jul 14, 2011

### Bill_K

gkangelexa, Studiot has already given you one example where what you're claiming is not true: melting, or more generally any change of state, requires an increase in energy but takes place at constant temperature. Even if you want to restrict yourself to gases, it doesn't hold for a Fermi gas (in which the kinetic energy remains nonzero even at T = 0) or a Bose gas like black-body radiation, in which u ∝ T4.

But even if you want to stay classical, it still doesn't hold! The internal energy of a monatomic gas is U = 3/2 kT, while for a diatomic gas is U = 5/2 kT. Now ask yourself: what happens to U in the transition region, in which a diatomic gas is dissociating into individual atoms?

7. Jul 14, 2011

### Philip Wood

Gkangelexa: I think you might have a serious misconception about the first law of thermodynamics. $\Delta$E = q + w doesn't mean that the internal energy has two components, q and w, or KE and PE. q and w refer to two ways in which energy can enter (or leave) the system: heat and work. These are not stored separately in the system.

An analogy I like is that of a bank account. E is your bank balance. q is inputs from cheques and w is inputs by electronic transfer. [Suppose there are no other methods of transfer.]