Question

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?

Answer 1

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