anonymous
  • anonymous
In a thermometer calibration test 0.31 OHM are registered when the temperature is 422 (K), according to R=R0*e^(A*((1/T)-(1/T0))) where A is a constant, R0 is in OHMs, and T0 in (K). If a R0=2.2 OHM for T0=310 (K), calculate R for 390 (°F).
Thermodynamics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
First, you need to find the value of A. Use the initial conditions along with the measured readings to find A. R=Ro*e^A*(1/T-1/To) so A = [ln(R/Ro)]/(1/T-1/To) = [ln(.31/2.2)]/(1/422-1/310) to find R for 390 F, convert F to Kelvin... 390 F = 472 K then used this value of T along with the value of A you calculated. so R=Ro*e^A*(1/T-1/To) =2.2*e^2289*(1/472-1/310)
anonymous
  • anonymous
thank you very much, I thought the first condition was R=R0*e^A*(1/T-1/422), the second R=0.22*e^A*(1/T-1/310) and the last R=Ro*e^A*(1/T-1/472), because only T0 is the registered temperature, I'll ask about both ideas, maybe the problem was not worded nicely

Looking for something else?

Not the answer you are looking for? Search for more explanations.