Recent content by johns123

Thanks Barryj .. I got it. I did the heat gained = heat lost Conservation of Energy ( heat ) m1C1(70 - T) + m2C2(70 - T) = m3C3( T - 5 ) with the m3 conversion going back to kcal/kg.C and I got a very reasonable answer. Amazing! Not one word of how to balance mixtures is anywhere in...

Homework Statement A glass mug has mass 125g empty. It contains 180g of coffee. Both are at 70 C. I add 15g of creme at 5 C. Find final temperature. Assume creme has specific heat of 2900 J/kg.C . Homework Equations Q = m C (delta)T in joules or Kcalories m = mass in kg...

I just reedited the problem above. Hope the errors are gone! My mistake. I meant m = 4.65 x 10^-23 grams/ion which I have to change to kg/ion. Also, I'm reading online at Yahoo that the problem should have specified a gas like Li where I could calculate Kg/mole of Li; go to kg/ion of Li...

Homework Statement The equation for the "average" velocity of a gas molecule in a closed container is: Vrms = ( 3 x k x T / m )^ 1/2 where k = 1.38 x 10 ^ 23 Joules/K T is temp in Kelvin m = mass per ion in kg = Atomic Mass M / Avagadro's Number Na so for N2, I...

Thanks Andrew. I think this one problem holds the key to a world of conversions that are left out of the books. I am using 3 books trying to learn one chapter.

I think I need to learn to type :-) it's 8.31 J/mk. I was just copying out of my notebook. Try again. n/V = 1.5 x 10^-9 kg/m^3 / .029 kg/mole = 51.724 x 10^-9 moles/m^3 and using p = (n/V)RT = 51.724 x 10^-9 ( 8.31 ) 500K = 214913.22 x 10^-9 pa ( I really hate this calculator ) thanks

Finally, I think I understand it ( I hope ) If air density at altitude is 1.5 x 10^-9 kg/m^3 at 500K and ( your hint ) M = .29 kg/mole that gives 1.5 x 10^-9 / .29 = 5.172 x 10^-9 moles/m^3 .. which is ( your hint ) = n/V and now use pV = nRT or p = (n/V)RT = 5.172 x 10^-9 ( 8.81...

OK. Thanks. I'm struggling with concepts here .. a battle of definitions. I'll solve it your way and post that.

Homework Statement At altitude of 160 km, the density of the air is 1.5 x 10^-9 kg/m^3 also the temp at that altitude is 500 K ( that's what it says! ) What is the pressure at that altitude? Homework Equations pV = nRT and mass density of air = "rho" the 1-values are at STP...

Yes. But when I needed to not ignore it, I didn't have a clue. It still seems a bit strange, but I finally realized that the gravitational force on the object was independent of it's volume and had to be considered separately. I might see this problem again with a heavy submarine .. and I'll be...

Finally, I see the concept: It's 2 problems in one. Net Force vertical .. and .. buoyant force vertical. Net Force = BF(air) - W(he) = W(air) - W(he) = p(air)*g*VolB - p(he)*g*VolB = MassB*Acc = p(he)*VolB*Acc then Acc = g * [ p(air) - p(he) ] / p(he) = 60.5 m/s^2 which is the correct...

Maybe something like BF = W(air) - W(he) = g * VolB( p(air) - p(he) ) = MassB * Acc then Acc = [ p(air) - p(he) / p(he) ] * g = [ (1.29 - .18) / .18 ] * 9.81 = 60.5 ! You sir, are a freakin' genius :-)

Homework Statement A small balloon filled with helium is released. What is it's initial upward acceleration? The only data given is ρ air = 1.29 kg/m^3 and ρ helium is .18 kg/m^3 I assume the buoyant force ( FB ) = the weight of the air displaced by the balloon so BF = W(air) = ρ(air)...

That's exactly what I said through the entire problem ? The overpressure is a calibrated pressure, and therefore the tire gage measures it wrong if used at other than 1 atm. If you notice in the problem, the professor only gives a way to calculate the tire gage error of .06 in the tire gage...

In other words, the original 2.4 measure is also in error, because the reading was taken at .95 atm. He gives that so the "overpressure error" of .06 can be calculated. I still find it difficult to determine the +/- sign of the calibration error. I do realize that the pressure inside the tire...