# 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?

Feb 1, 2017

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

$\text{Rate of heat flow}$
$= \text{Thermal Conductivity" × "Temp Difference" × "Area" / "Length}$
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$= m \times L$
where $L$ is latent heat of fusion of water and is$= 79.7 c a l \cdot {g}^{-} 1$
$Q = 10 \times 79.7 = 797 c a l$
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
$\implies \dot{Q} = 2.59 c a l \cdot {s}^{-} 1$
3. Number of seconds required to conduct $797 c a l$ of heat$= \frac{797}{2.59} = 308 s$, rounded to last second