91影视

1. Length of the copper rod is$L=1.2\text{鈥�}\text{m}.$
2. Cross sectional area of the copper rod is$A=4.8{\text{cm}}^{\text{2}}\text{or}4.8脳{10}^{鈭�4}{\text{m}}^{\text{2}}.$
3. The temperature difference between the two ends of the rod is$螖T=100\text{掳颁鈥夆赌塷谤100碍}$
4. Thermal conductivity of copper is.$k_c=\text{鈥�}401\text{鈥夆赌�}\frac{\text{W}}{\text{m.K}}$

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, $P_{cond}=\frac{kA\left(螖T\right)}{L}$ 鈥�(颈)

Where,

K is the thermal conductivity

A is the cross sectional area

$螖T$is the change in temperature

L is length of the copper rod

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

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

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

$\begin{array}{c}{P}_{\text{cond}}=\frac{401\text{鈥�}\frac{\text{W}}{\text{m.K}}(4.8脳{10}^{鈭�4}{\text{鈥尘}}^{\text{2}})\left(100\text{鈥夆赌塊}\right)}{1.2\text{鈥尘}}\\ =16\text{鈥塉/蝉}\end{array}$

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

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

$\begin{array}{c}\frac{dm}{dt}=\frac{16\text{鈥夆赌塉/蝉}}{33600\text{J/kg}}\\ =48\text{鈥夆赌塳g/s}\\ =0.048\text{鈥塯/蝉}\end{array}$

Therefore, the rate at which ice melts is $0.048\text{鈥塯/蝉}$.