How to find maximum temp. rise from a velocity value

[pondwire]

Homework Statement

Wind (dry air, so R=287 Jkg^-1K^-1) is moving isobarically along a flat surface at 30m/s. It hits a forest and 3/4 of its kinetic energy is transformed into heat energy by friction. What is the maximum temperature increase that could occur?

Homework Equations

The Attempt at a Solution

Using the potential temperature equation, theta = (1000/p)^(R/Csub p), and this equation: changeT = pot. temp. - initial temp., I get a change in temp. of -0.004T, which does nothing for me because there is no velocity included. I also tried using kinetic energy (1/2mv^2) to try to relate it to work and then internal energy, but I don't have a mass for the air parcel.

 

[rock.freak667]

Well I am not sure if I am doing this correctly but

you can find the heat in terms of mass which is 3/4E_k. This is the should be the same as work done = pV.

And the ideal gas equation with the specific gas constant is pV=mRT.

 

[pondwire]

Thank you for the reply, but I still can't get rid of the mass m in kinetic energy. I've tried using Work = change in KE = - change in potential energy and STILL this mass m is there. Can anyone help?

 

[Delphi51]

I also tried using kinetic energy (1/2mv^2) to try to relate it to work and then internal energy, but I don't have a mass for the air parcel.

I think you are on the right track here. You could just say you are doing it for each kg of air. Or cancel the m's in .5*m*v^2 = m*C*(delta T) .

 

[pondwire]

Hey, thanks both you guys for trying to help. I DID get rid of mass m when I set mCchangeT equal to 1/2mv^2, but still did not find the right answer (the book actually gives the answer of maximum temperature change of 0.15 Kelvin, but not much help in showing how to get there.) It seems like it should be so simple! If anyone else out there has any other ideas, I'm listening...

 

[Delphi51]

Interesting. I used C = 1005 J/kg/degree and got a temperature rise of 0.336 degrees.
It seems like a straightforward problem, but I guess not. What is this R quantity? It has units of heat capacity but not the standard heat capacity for air.