One end of a cylindrical copper rod 1.2 m long with a 2 cm diameter is in boiling water at 100°C. The other end is in the ice at 0°C. b. How much time in seconds is required to melt 10 grams of ice?

1 Answer
Feb 1, 2017

Basic equation connecting thermal conductivity extrinsic factors like heat flow, temperature gradient area and length is

#"Rate of heat flow" #
#= "Thermal Conductivity" × "Temp Difference" × "Area" / "Length"#
#dot Q = k × ∆T × A/L # ......(1)
Note that #dot Q# is in Watts or Calory per second; it is power, or energy per unit time.

  1. Heat #Q# required to melt #10# grams of ice#=mxxL#
    where #L# is latent heat of fusion of water and is#=79.7cal cdot g^-1#
    #Q=10xx79.7=797cal#
  2. Thermal conductivity of Copper #k=0.99 calcdot s^-1cm^-1·K^-1#
    Inserting given values in equation (1) we get
    #dot Q = 0.99 × (100-0) × (pixx1^2)/120 #
    #=>dot Q = 2.59calcdot s^-1 #
  3. Number of seconds required to conduct #797cal# of heat#=797/2.59=308s#, rounded to last second