A cylindrical copper rod of length$\mathbf{1}\mathbf{.2}\mathbf{}\mathbf{\text{m}}$and cross sectional area$\mathbf{4}\mathbf{.8}\mathbf{}{\mathbf{\text{\hspace{0.17em}cm}}}^{\mathbf{\text{2}}}$is insulated to prevent heat loss through its surface. The ends are maintained at a temperature difference of$\mathbf{100}{}^{\mathbf{\text{0}}}\mathbf{\text{C}}$by having one end in a water – ice mixture and the other in a mixture of boiling water and steam.(a) At what rate is energy conducted along the rod? (b) At what rate does ice melt at the cold end?

1. Length of the copper rod is$L=1.2\text{m}.$
2. Cross sectional area of the copper rod is$A=4.8{\text{cm}}^{\text{2}}\text{or}4.8\times {10}^{-4}{\text{m}}^{\text{2}}.$
3. The temperature difference between the two ends of the rod is$\Delta T=100\text{°C\hspace{0.17em}\hspace{0.17em}or100K}$
4. Thermal conductivity of copper is.$k_c=401\frac{\text{W}}{\text{m.K}}$

Conduction, or any other mode of heat transfer, can only happen when the temperatures of the two materials are different. We can find the rate at which energy is conducted by the rod by using the formula for it. We can get the rate at which ice melts from the conduction rate and latent heat of fusion.

Formula:

The rate of conduction of the energy, $P_{cond}=\frac{kA\left(\Delta T\right)}{L}$ …(i)

Where,

K is the thermal conductivity

A is the cross sectional area

$\Delta T$is the change in temperature

L is length of the copper rod

The rate of melting of a body due to conduction,$\frac{dm}{dt}=\frac{P_{\text{cond}}}{L_F}$ …(ii)

$L_F$is the latent heat of fusion of ice = $33600\text{J/kg}$

Conduction rate of the energy using equation (i) is given as:

$\begin{array}{c}{P}_{\text{cond}}=\frac{401\frac{\text{W}}{\text{m.K}}(4.8\times {10}^{-4}{\text{\hspace{0.17em}m}}^{\text{2}})\left(100\text{\hspace{0.17em}\hspace{0.17em}K}\right)}{1.2\text{\hspace{0.17em}m}}\\ =16\text{\hspace{0.17em}J/s}\end{array}$

Therefore, the rate at which energy is conducted by the rod is.$16\text{J/s}$

The rate at which ice melts is given using equation (ii) as follows:

$\begin{array}{c}\frac{dm}{dt}=\frac{16\text{\hspace{0.17em}\hspace{0.17em}J/s}}{33600\text{J/kg}}\\ =48\text{\hspace{0.17em}\hspace{0.17em}kg/s}\\ =0.048\text{\hspace{0.17em}g/s}\end{array}$

Therefore, the rate at which ice melts is $0.048\text{\hspace{0.17em}g/s}$.