Question

A conducting rod of length $1\ \text{m}$ has area of cross-section $10^{-3}\ \text{m}^2$. One end is immersed in boiling water ($100^\circ\text{C}$) and the other end in ice ($0^\circ\text{C}$). If coefficient of thermal conductivity of copper is $92\ \text{cal}/\text{m-s-}^\circ\text{C}$ and latent heat of ice is $8 \times 10^4\ \text{cal}/\text{kg}$, then the amount of ice which will melt in one minute is

• A. $5.4 \times 10^{-3}\ \text{kg}$
• B. $6.9 \times 10^{-3}\ \text{kg}$
• C. $1.8 \times 10^{-3}\ \text{kg}$
• D. $3.6 \times 10^{-3}\ \text{kg}$

Hint:

Always ensure your time units are consistent. The thermal conductivity is given per second, so the 1 minute duration must be strictly converted to 60 seconds before calculating the total heat transferred.

Answer 1

The Correct Option is B

Step 1: Picture the setup.
Heat travels along a copper rod from boiling water ($100^\circ$C) at one end to ice ($0^\circ$C) at the other. That heat melts some ice. We find how much melts in one minute.

Step 2: Recall the conduction rule.
Heat per second through a rod is \[ \frac{Q}{t}=\frac{KA\,\Delta T}{L}, \] where $K$ is conductivity, $A$ area, $\Delta T$ the temperature difference, and $L$ the length.

Step 3: List the values.
$K=92$, $A=10^{-3}\ \text{m}^2$, $\Delta T=100-0=100$, $L=1$ m, $t=60$ s, latent heat $L_f=8\times10^4\ \text{cal/kg}$.

Step 4: Find the total heat in one minute.
\[ Q=\frac{KA\,\Delta T}{L}\times t=\frac{92\times10^{-3}\times100\times60}{1}. \]

Step 5: Do the arithmetic.
\[ Q=9.2\times60=552\ \text{cal}. \]

Step 6: Link heat to melted mass.
Melting needs $Q=mL_f$, so \[ m=\frac{Q}{L_f}=\frac{552}{8\times10^4}=69\times10^{-4}\ \text{kg}. \]

Step 7: State the result.
\[ m=6.9\times10^{-3}\ \text{kg}, \] which is option (2).
\[ \boxed{6.9\times10^{-3}\ \text{kg}} \]