1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Please i cannot find any help on this question so far

  1. Apr 14, 2008 #1
    please... i cannot find any help on this question so far!!

    1. The problem statement, all variables and given/known data
    A rod is initially at a uniform temperature of 0 ^deg C throughout. One end is kept at 0 ^deg C, and the other is brought into contact with a steam bath at 100 ^deg C. The surface of the rod is insulated so that heat can flow only lengthwise along the rod. The cross-sectional area of the rod is 2.50 cm^2, its length is 120 cm, its thermal conductivity is 380 W/m * K, its density is 1.00 x 10^4 kg/m^3, and its specific heat capacity is 520 J/ kg * K. Consider a short cylindrical element of the rod 1.00 cm in length.

    1. If the temperature gradient at the cooler end of this element is 140 deg C/m, how many joules of heat energy flow across this end per second?

    2. If the average temperature of the element is increasing at the rate of 0.250 deg C/sec, what is the temperature gradient at the other end of the element?

    2. Relevant equations

    3. The attempt at a solution
    I have gotten number 1 to be 13.3W, but do not know how to proceed with number 2. Any help would be greatly appreciated!
  2. jcsd
  3. Apr 15, 2008 #2
    both 1 & 2 are about a short cylindrical element of the rod 1.00 cm in length.

    You know the heat going out at the cooler end. You know how quickly the temperature rises and you can calculate how much the internal energy of the element increases, so you should be able to calculate how much heat energy goes in at the hot end.
  4. Apr 15, 2008 #3
    I am still stuggling. I see that from the equation H = kA(Th-Tc)/L, i can say that H is equal to dQ/dt, which is the rate of the heat current. I took this information and applied my .25 celsius degrees to the equation.

    .25/(kA) = (Th-Tc)/L ---the temp. gradient

    This is wrong though, what am I missing. I am desperate at this point!!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?