Thermodynamics: Conservation of energy problem

[lwelch70]

Homework Statement

A closed system initially at rest on the surface of the Earth undergoes a process for which there is a net energy transfer to the system by work of 200 Btu. During the process, there is a net heat transfer from the system of 30 Btu. At the end of the process, the system has velocity of 200 ft/sec at an elevation of 200 ft. The mass of the system is 50lb and the local acceleration due to gravity is 32 ft/s². Determine the change in internal energy of the system for this process, in Btu.

Homework Equations

\DeltaE = \DeltaU + \DeltaKE + \Delta PE

The Attempt at a Solution

So I understand that the system initially has 200 Btu transferred to it. And 30 Btu taken away from the heat.

So my understanding of that is that \DeltaE is equal to 170 Btu. Correct?

In calculating the \DeltaKE, all I see is that initially it is 0, and after it is 1/2mv². This gives my units in \frac{lbft^2}{s^2}

How do I get this in Btu? I'm not very good with units.

Also, would I include \DeltaPE since the system is moving at the end of process?

 

[lwelch70]

Anyone?

 

[lwelch70]

No help?

 

[hcelik]

Maybe this can help you;

Btu / lbm = 25,037 ft2 / s2

or you can also use any conversation software/website from pound force to btu.

 

[rude man]

I think everything you've written is good. Yes, the increase in potential energy must be included in your equation.

I also don't like dealing in the British units system. It doesn't help that there are more than one of them - BG and EE and god knows what else (see link below). The thing you need is converting BTU to ft-pounds: •1 Btu (British thermal unit) = 778.3 ft-lb. Now everything is in pounds of mass, pounds of force, feet and seconds. In this case I suppose we're dealing with the EE system since mass is given in pounds, not slugs. So 1 lb of force = 1 lb of mass * 32.2 ft/sec2. The last number is of course g, the acceleration of gravity, which everybody but the Brits know as 9.81 m/sec2. :-)

Actually, there IS a similar confusion with folks using kg - they use kg to mean both force and mass, exactly analogous to the misuse of pounds. Scientists of course avoid using kg for force or weight, using Newtons instead.


http://www.engineeringtoolbox.com/mass-weight-d_589.html