Question

A copper rod has a length of 1.3 m and a cross-sectional area of 3.6 10-4 m2. One end of the rod is in contact with boiling water and the other with a mixture of ice and water. What is the mass of ice per second that melts? Assume that no heat is lost through the side surface of the rod.

Answer 1

To solve the problem it is necessary to apply the concepts related to heat flow,

The heat flux can be defined as

`(dQ)/(dt) = H = (kA\Delta T)/(d)`

Where,

k = Thermal conductivity

A = Area of cross-sectional area

d = Length of the rod

`\Delta T=` Temperature difference between the ends of the rod

`k =388 W/m.\°C` Thermal conductivity of copper rod

`A = 3.6 *10^(-4) m` Area of cross section of rod

`\Delta T=100-0=100\°C` Temperature difference

`d=1.3m` length of rod

Replacing then,

`H = (kA\Delta T)/(d)`

`H = ((388)(3.6 *10^(-4))(100))/(1.3)`

`H=10.7446J`

From the definition of heat flow we know that this is also equivalent

`H = \dot{m}*L`

Where,

`\dot{m} =` Mass per second

`L = 334J/g` Latent heat of fusion of ice

Re-arrange to find
`\dot{m},`

`H = \dot{m}*L`

`\dot{m}=(L)/(H)`

`\dot{m}=(334)/(10.7446)`

`\dot{m} = 31.08g/s`

`\dot{m}= 0.032g/s`

Therefore the mass of ice per second that melts is 0.032g