Question

One end of a copper rod of length $2 m$ and area of cross section $10^{-3} m ^{2}$ is immersed in boiling water and the other in solid form of water (ice). If the coefficient of thermal conductivity of copper is $92 cal / m - s -{ }^{\circ} C$ and the latent heat of ice is $9 \times 10^{4}$ cal/kg (assumed) then the amount of ice which will melt in three minute is

• A. $15 \times 10^{-1} kg$
• B. $40 \times 10^{-2} kg$
• C. $92 \times 10^{-4} kg$
• D. $25 \times 10^{-2} kg$

Answer 1

Answer: (C)

$mL =\frac{ KA (\Delta T ) t }{ x } $
$\Rightarrow m =\frac{92 \times 10^{-3} \times 100 \times 180}{2 \times 9 \times 10^{4}}$
$m =92 \times 10^{-4} kg$