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

$\frac{dQ}{dt} = H = \frac{kA\Delta T}{d}$

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,

$H = \frac{kA\Delta T}{d}$

$H = \frac{(388)(3.6 *10^{-4})(100)}{1.3}$

$H=10.7446J$

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 $\dot{m},$

$H = \dot{m}*L$

$\dot{m}=\frac{L}{H}$

$\dot{m}=\frac{334}{10.7446}$

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

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

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