# Equation for Internal Energy?

Tags:
1. Apr 28, 2017

### SpiraRoam

1. The problem statement, all variables and given/known data
Sketch a diagram of internal energy (y-axis) versus temperature in the range from -10°C to +112°C to indicate how energy would change for a fixed quantity of water in its three phrases. Label and explain the main features of the variation.

2. Relevant equations

3. The attempt at a solution
I'm going to show how energy would hange over the 3 phases of freezing whilst a solid, melting from a solid and vapourising from a liquid aswell as the triple point.

Thing is I'm not sure whether the first equation mc delta theta is the right one for internal energy alone? Would theta represent the change in absolute temperature from -273 degrees or the particles? I assume it's mass x speed of light x the change in absolute temp or something to do with the particles?

Or is another equation entirely?

Cheers

2. Apr 28, 2017

### Staff: Mentor

C in the equation is not the speed of light. Guess again.

3. Apr 28, 2017

### SpiraRoam

specific or just normal heat capacity?

4. Apr 28, 2017

### Staff: Mentor

Specific

5. Apr 28, 2017

### SpiraRoam

Okay, cheers. So that is definitely the equation used to find the internal energy of a quantity of water in whatever state? It doesn't change for a gas? There's and it's relatives in my notes as well and I've got a feeling they may be relevant to gases.

Is theta the change in absolute temperature then?

6. Apr 28, 2017

### Staff: Mentor

No. $\Delta \theta$ is the change in temperature.

You really need to do some studying in thermo book before you start working on this question. You are lacking the required background.

7. Apr 28, 2017

### SpiraRoam

Yeah you're right, I only gave it a general overview. Absolute temp is the difference from -273 degrees isn't it. So mass x specific heat capacity x change in temperature. I just need to find out how the mass changes in phases and states so I can calculate and begin to plot.

The thing is - a mass is only given in part B of the question (Specific capacity of water 4.19 kJ kg–1 K–1) and it isn't specified whether that applies to part A aswell...but it must do for the equation to work. How else am I going to get any values in Joules?

8. Apr 28, 2017

### Staff: Mentor

They only asked for a (rough) sketch.

9. Apr 28, 2017

### SpiraRoam

I hope so! It's hard to know whether to take the phrasing literally or not with it being science XD