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

$\Delta$E = $\Delta$U + $\Delta$KE + $\Delta$ 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 $\Delta$E is equal to 170 Btu. Correct?

In calculating the $\Delta$KE, 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 $\Delta$PE 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