Acurrent of 300 a is passed through a stainless-steel wire [k 19 w/m* c] 3 mm in diameter. the resistivity of the steel may be taken as 70 4 cm, and the length of the wire is 1 m. the wire is submerged in a liquid at 110 oc and experiences a convection heat-transfer coefficient of 4 kw/m2 c a) calculate the center temperature of the wire? b) what is the temp 25mm from the otter surface of the wire? c) the melting point for the stainless wire is 1363 .c. what is the wire before it begins to melt? max current that can be applied to the all the power generated in the wire must be dissipated by convection to the liquid: p=2r-q-h4 (tw-too) equation (a) the resistance of the wire is calculated from r-p l/a= (70x10-6)(100)/(0.15)2-0.099 ohms where p is the resistivity of the wire. the surface area of the wire is adl, so from equation (a), (200)2(0.099)4000x(3x10-3)(1)(tw-11 0)-3960 w & tw-215 c the heat generated per unit volume q is calculated from p=qv= qur2l q 3960/t (1.5x10-3)2(1) =560.2 mw/m3 where p is the resistivity of the wire. the surface area of the wire is tdl, so from equation (a), (200)2(0.099)-4000x(3x10-3)(1)(tw-1 10)-3960 w 1s

  remember that hf is only the energy of one photon. not two. not three. one.  1) 5.12 * 10^3 photons  find the energy of one photon, then determine how many photons are needed to reach the sensitivity.  e = hf  = (6.626 * 10^-34 js)(5.95 * 10^19 hz)  = 3.94247 * 10^-14 j per photon  (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons  round to three sig figs.  2) 5.77 * 10^-17 j  energy of pulse = energy per photon * number of photons  = hfn  = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250)  = 5.774559 * 10^-17 j  round to three sig figs.  3) 4.07 * 10^8 kj/mol  10^12 pm = 1 m  c = fλ  f = c/λ  e = hf  = hc/λ  = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol)  = 40702571.4 kj/mol  round to three sig figs.  4) 9.94 * 10^-1 m  e = hc/λ  λ = hc / e  = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j)  = 0.9939 m  round to three sig figs.  5) 9.06 * 10^3 photons  energy of pulse = energy per photon * number of photons  number of photons = energy of pulse / energy per photon  = energy of pulse / (hf)  = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)]  = 9055.23695 photons  round to three sig figs.  6) 2.78 * 10^-15 j  this is problem 1, except in reverse.  sensitivity = minumum number of photons * energy per photon  = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz)  = 2.78070758 × 10^-15 j  round to three sig figs.

Ithink its called the coefficent.