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Thermal Conductivity and latent heat

  1. Jun 15, 2014 #1
    1. The problem statement, all variables and given/known data

    If a copper kettle has a base of thickness 2.0mm and an area 3.0 x 10-2 m2 estimate the steady difference in temperature between the inner and outer surface of the base which
    must be maintained to enable enough heat too pass through so that the temperature of 1 kg of water
    rises of 0.25 K/s assume that there are no heat losses.

    (ii) After reaching the temperature of 373K the water is allowed to boil under the same conditions for 120 seconds and the mass of water remaining in the kettle is 0.948kg.
    Deduce a value for the specific latent heat of vaporization of water ( neglecting condensation of the steam in the kettle)

    2. Relevant equations

    Thermal conductivity of copper 3.8 x 102 W/m/C
    Specific Heat Capacity of water 4.0 x 103 J/kg/K

    Equations used

    dQ/dt = kA(dΘ/dx)

    where dQ/dt = rate of flow of heat
    K = Thermal Conductivity
    A = Area
    dΘ = Change in Temperature
    dx = Thickness
    whereby dΘ/dx = Temperature gradient

    Q = mCT

    Where Q = Heat require
    m = Mass of substance
    C = Specific heat capacity
    T = Temperature

    Q= mL

    Where Q = Heat Require
    m = Mass
    L = Latent Heat

    3. The attempt at a solution

    Firstly i used the specific heat capacity equation to find the rate of flow of heat

    Q = 1 (4.2 x 103 J/kg/K ) 0.25 K/s
    = 1050 J/s

    Using the above answer i substitute it into the equation for thermal conductivity

    1050 = 3.8 x 102 W/m/C (3.0 x 10-2 m2 )(dΘ/0.002m)
    by simple calculations my answer = 0.18 degrees

    Part (ii)

    I am abit lost as to where to go from here... but im thinking i will have to multiply 120 seconds by 0.25K/s which = 30 K/s

    and used the 30K/s in the specific heat capacity equation ...( but im thinking that doesnt make any sense)

    By using the 373k and substituting it into the heat capacity equation

    Q = 1(4.2 x 103 J/kg/K)373
    = 1566600J

    and using the answer above in the Latent heat of vaporization equation

    1566600 = 0.948L
    L = 1.6 x 106 J/kg/K

    This seems wrong tho...

    By the way is my answer correct for Part (i) ??
     
    Last edited: Jun 15, 2014
  2. jcsd
  3. Jun 15, 2014 #2

    DrClaude

    User Avatar

    Staff: Mentor

    The temperature will be constant, so it's abit strange to start with something in K/s!

    In part (i), you figured out how much power is furnished to the kettle. Start from there.

    I didn't check all the calculations, but the approach seemed right.
     
  4. Jun 15, 2014 #3
    :biggrin: YAYYYYY i got!!!.....geezeeee that was simple enough....

    ok this is what i did ...please let me kno if im correct...although i think i am

    soo you said to start from the power i got in part (i) which was 1050 J/s

    Therefore for rate of flow of heat( which is power) dQ/dt = 1050watts

    since the time was given in the problem, 120seconds, i substituted it in for dt

    solving for dQ i got 12600

    then by subtracting the initial mass from the final, i got the mass of water that was vaporized
    = 0.052

    these values was then substituted into the formula for latent heat of vaporization

    giving me 2.4 x 106
     
  5. Jun 15, 2014 #4

    DrClaude

    User Avatar

    Staff: Mentor

    Great, but don't forget the units :tongue2:
     
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