A steel rod ( K = 46 J s^ -1 m^ -1 ^ ^ ^ ^ -1 ) of length 1.0 m has...

The rate of heat transfer through the rod is given by Fourier's law of heat conduction:

$H = \dfrac{KA(T_1 - T_2)}{L}$

Substituting the given values:
K = 46 \text{ J s}^{-1} \text{ m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}
$A = 0.04 \text{ cm}^2 = 4 \times 10^{-6} \text{ m}^2$
T_1 - T_2 = 100^{}^{{}^{\circ}}\text{C} - 0^{}^{{}^{\circ}}\text{C} = 100^{}^{{}^{\circ}}\text{C}
$L = 1.0 \text{ m}$

$H = \dfrac{46 \times 4 \times 10^{-6} \times 100}{1.0} = 1.84 \times 10^{-2} \text{ J s}^{-1}$

The heat transferred per second is used to melt the ice. The mass of ice melting per second $m$ is given by:

$m = \dfrac{H}{L_f}$

Given the latent heat of fusion of ice $L_f = 3.36 \times 10^5 \text{ J kg}^{-1}$:

$m = \dfrac{1.84 \times 10^{-2}}{3.36 \times 10^5} \text{ kg s}^{-1}$

$m \approx 5.476 \times 10^{-8} \text{ kg s}^{-1}$

Converting the mass to grams per second:

$m = 5.476 \times 10^{-8} \times 10^3 \text{ g s}^{-1} \approx 5.5 \times 10^{-5} \text{ g s}^{-1}$