Question

A copper rod of cross-sectional area 11.6 cm2 has one end immersed in boiling water and the other in an ice-water mixture, which is thermally well insulated except for its contact with the copper. The length of the rod between the containers is 19.6 cm, and the rod is covered with a thermal insulator to prevent heat loss from the sides.

How many grams of ice melt each second? (The thermal conductivity of copper is 390 W/m-C°.)

Answer 1

0.686 g of ice melts each second.

Solution:

As per the question:

Cross-sectional Area of the Copper Rod, A =
`11.6\ cm^(2) = 11.6* 10^(- 4)\ m^(2)`

Length of the rod, L = 19.6 cm = 0.196 m

Thermal conductivity of Copper, K =
`390\ W/m.^(\circ)C`

Conduction of heat from the rod per second is given by:

`q = (KA\Delta T)/(L)`

where

`\Delta T = 100^(\circ) - 0^(\circ) = 100^(\circ)C` = temperature difference between the two ends of the rod.

Thus

`q = (390* 11.6* 10^(- 4)* 100)/(0.196) = 228.48\ J/s`

Now,

To calculate the mass, M of the ice melted per sec:

`M = (q)/(L_(w))`

where

`L_(w)` = Latent heat of fusion of water = 333 kJ/kg

`M = (228.48)/(333* 10^(3)) = 6.86* 10^(- 4)\ kg = 0.686\ g`